23
Electric Fields
CHAPTER OUTLINE
23.1
23.2
23.3
23.4
23.5
23.6
23.7
Properties of Electric
Charges
Charging Objects by
Induction
Coulomb’s Law
The Electric Field
Electric Field of a
Continuous Charge Distribution
Electric Field Lines
Motion of a Charged Particle
in a Uniform Electric Field
ANSWERS TO QUESTIONS
Q23.1
A neutral atom is one that has no net charge. This
means that it has the same number of electrons
orbiting the nucleus as it has protons in the nucleus.
A negatively charged atom has one or more excess
electrons.
*Q23.2
(i) Suppose the positive charge has the large value 1 mC.
The object has lost some of its conduction electrons,
in number 10−6 C (1 e/1.60 × 10−19 C) = 6.25 × 1012 and
in mass 6.25 × 1012 (9.11 × 10−31 kg) = 5.69 × 10−18 kg.
This is on the order of 1014 times smaller than the
~1g mass of the coin, so it is an immeasurably small
change. Answer (d).
(ii) The coin gains extra electrons, gaining mass on
the order of 10−14 times its original mass for the charge
−1 m C. Answer (b).
Q23.3
All of the constituents of air are nonpolar except for water. The polar water molecules in the
air quite readily “steal” charge from a charged object, as any physics teacher trying to perform
electrostatics demonstrations in the summer well knows. As a result—it is difficult to accumulate
large amounts of excess charge on an object in a humid climate. During a North American
winter, the cold, dry air allows accumulation of significant excess charge, giving the potential
(pun intended) for a shocking (pun also intended) introduction to static electricity sparks.
Q23.4
Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses)
of two particles, and inversely proportional to the square of the separation distance. An electrical
force exhibits the same proportionalities, with charge as the intrinsic property.
Differences: The electrical force can either attract or repel, while the gravitational force as
described by Newton’s law can only attract. The electrical force between elementary particles is
vastly stronger than the gravitational force.
Q23.5
No. The balloon induces polarization of the molecules in the wall, so that a layer of positive
charge exists near the balloon. This is just like the situation in Figure 23.4a, except that the signs
of the charges are reversed. The attraction between these charges and the negative charges on the
balloon is stronger than the repulsion between the negative charges on the balloon and the
negative charges in the polarized molecules (because they are farther from the balloon), so that
there is a net attractive force toward the wall. Ionization processes in the air surrounding the
balloon provide ions to which excess electrons in the balloon can transfer, reducing the charge
on the balloon and eventually causing the attractive force to be insufficient to support the weight
of the balloon.
1
2
Chapter 23
*Q23.6 Answer (c). Each charge produces field as if it were alone in the Universe.
*Q23.7 (i) According to the inverse square law, the field is one-fourth as large at twice the distance. The
answer is (c), 2 × 36 cm = 72 cm.
(ii) The field is four times stronger at half the distance away from the charge. Answer (b).
Q23.8
An electric field created by a positive or negative charge extends in all directions from the charge.
Thus, it exists in empty space if that is what surrounds the charge. There is no material at point A
in Figure 23.21(a), so there is no charge, nor is there a force. There would be a force if a charge
were present at point A, however. A field does exist at point A.
*Q23.9 (i) We compute qAqB/r2 in each case. In (a) it is 400/4 = 100 (nC/cm)2. In (b) and (c), 300/4 =
75 (nC/cm)2. In (d) 600/9 = 67 (nC/cm)2. In (e) 900/9 = 100 (nC/cm)2. The ranking is then a = e >
b = c > d.
(ii) We compute qA/r2 in each case. In (a) it is 20/4 = 5 nC/cm2. In (b) 30/4 = 7.5 nC/cm2. In (c)
10/4 = 2.5 nC/cm2. In (d) 30/9 = 3.3 nC/cm2. In (e) 45/9 = 5 nC/cm2. The ranking is then b > a =
e > d > c.
*Q23.10 The charge at the upper left creates at the field point electric field to the left, with magnitude we
call E1. The charge at lower right creates downward electric field with an equal magnitude E1.
These two charges together create field 2 E1 downward and to the left at 45°. The positive
charge is 2 times farther from the field point so it creates field 2E1/( 2 )2 = E1 upward and
to the right. The net field is then 2 − 1 E1 downward and to the left. The answer to question
(i) is (d).
(
)
(ii) With the positive charge removed, the magnitude of the field becomes 2 E1, larger than
before, so the answer is (a).
*Q23.11 The certain point must be on the same line as A and B, for otherwise the field components
perpendicular to this line would not add to zero. If the certain point is between A and B, it is
midway between them, and B’s charge is also +40 nC. If the certain point is 4 cm from A
and 12 cm from B, then B’s charge must be −9(40 nC) = −360 nC. These are the only two
possibilities. The answers are (a), (f), and (j).
Q23.12 The direction of the electric field is the direction in which a positive test charge would feel a force
when placed in the field. A charge will not experience two electrical forces at the same time, but
the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at
which they cross would feel a force in two directions. Furthermore, the path that the test charge
would follow if released at the point where the field lines cross would be indeterminate.
Q23.13 Both figures are drawn correctly. E1 and E2 are the electric fields separately created by the point
charges q1 and q2 in Figure 23.12 or q and –q in Figure 23.13, respectively. The net electric field
is the vector sum of E1 and E2 , shown as E. Figure 23.19 shows only one electric field line at
each point away from the charge. At the point location of an object modeled as a point charge, the
direction of the field is undefined, and so is its magnitude.
*Q23.14 Answer (a). The equal-magnitude radially directed field contributions add to zero.
*Q23.15 Answer (c). Contributions to the total field from bits of charge in the disk lie closer together in
direction than for the ring.
Electric Fields
3
*Q23.16 (i) Answer (c). Electron and proton have equal-magnitude charges.
(ii) Answer (b). The proton’s mass is 1836 times larger than the electron’s.
*Q23.17 Answer (b).
Q23.18 Linear charge density, l, is charge per unit length. It is used when trying to determine the
electric field created by a charged rod.
Surface charge density, s, is charge per unit area. It is used when determining the electric field
above a charged sheet or disk.
Volume charge density, r, is charge per unit volume. It is used when determining the electric field
due to a uniformly charged sphere made of insulating material.
Q23.19 No. Life would be no different if electrons were + charged and protons were – charged. Opposite
charges would still attract, and like charges would repel. The naming of + and – charge is merely
a convention.
Q23.20 In special orientations the force between two dipoles can be zero or a force of repulsion. In
general each dipole will exert a torque on the other, tending to align its axis with the field created
by the first dipole. After this alignment, each dipole exerts a force of attraction on the other.
SOLUTIONS TO PROBLEMS
Section 23.1
P23.1
(a)
Properties of Electric Charges
The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its
mass by a negligible amount, to
1.007 9 (1.660 × 10 −27 kg ) − 9.11 × 10 −31 kg = 1.67 × 10 −27 kg .
Its charge, due to loss of one electron, is
0 − 1( −1.60 × 10 −19 C ) = +1.60 × 10 −19 C .
(b)
By similar logic, charge = +1.60 × 10 −19 C
)
mass = 22.99 (1.66 × 10 −27 kg − 9.11 × 10 −31 kg = 3.82 × 10 −26 kg
(c)
charge of Cl− = −1.60 × 10 −19 C
)
−27
−31
−26
mass = 35.453(1.66 × 10 kg + 9.11 × 10 kg = 5.89 × 10 kg
(d)
)
charge of Ca ++ = −2 ( −1.60 × 10 −19 C = +3.20 × 10 −19 C
)
)
−27
−31
−26
mass = 40.078 (1.66 × 10 kg − 2 ( 9.11 × 10 kg = 6.65 × 10 kg
(e)
)
kg ) + 3( 9.11 × 10
charge of N 3− = 3( −1.60 × 10 −19 C = −4.80 × 10 −19 C
mass = 14.007 (1.66 × 10 −27
−31
)
kg = 2.33 × 10 −26 kg
continued on next page
4
Chapter 23
(f)
)
charge of N 4 + = 4 (1.60 × 10 −19 C = +6.40 × 10 −19 C
mass = 14.007 (1.66 × 10 −27 kg ) − 4 ( 9.11 × 10 −31 kg ) = 2.32 × 10 −26 kg
(g)
We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.
)
charge = 7 (1.60 × 10 −19 C = 1.12 × 10 −18 C
)
)
mass = 14.007 (1.66 × 10 −27 kg − 7 ( 9.11 × 10 −31 kg = 2.32 × 10 −26 kg
(h)
charge = −1.60 × 10 −19 C
mass = ⎡⎣ 2 (1.007 9 ) + 15.999 ⎤⎦ 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 2.99 × 10 −26 kg
P23.2
(a)
⎛
⎞⎛
10.0 grams
electrons ⎞
23 atoms ⎞ ⎛
24
N =⎜
⎟ = 2.62 × 10
⎟ ⎜ 47
⎜ 6.02 × 10
mol ⎠ ⎝
atom ⎠
⎝ 107.87 grams mol ⎟⎠ ⎝
(b)
# electrons added =
or
Q
1.00 × 10 −3 C
= 6.25 × 1015
=
e 1.60 × 10 −19 C electron
2.38 electrons for every10 9 already present .
Section 23.2
Charging Objects by Induction
Section 23.3
Coulomb’s Law
P23.3
If each person has a mass of ≈70 kg and is (almost) composed of water, then each person
contains
⎛ 70 000 grams ⎞ ⎛
molecules ⎞ ⎛
protons ⎞
6.02 × 10 23
N ≅⎜
10
≅ 2.3 × 10 28 protons
⎟
⎝
⎠
⎝
⎝ 18 grams mol ⎠
mol
molecule ⎠
With an excess of 1% electrons over protons, each person has a charge
q = 0.01(1.6 × 10 −19 C ) ( 2.3 × 10 28 ) = 3.7 × 10 7 C
ª
So
F = ke
(3.7 × 10
q1q2
= ( 9 × 10 9 )
2
r
0.62
)
7 2
N = 4 × 10 25 N ~10 26 N
This force is almost enough to lift a weight equal to that of the Earth:
Mg = 6 × 10 24 kg ( 9.8 m s 2 ) = 6 × 10 25 N ~ 10 26 N
Electric Fields
*P23.4
k q q
In the first situation, FA on B,1 = e A2 B ˆi . In the second situation, qA and qB are the same.
r1
k q q
FB on A,2 = − FA on B = e A2 B − ˆi
r2
( )
F2 ke qA qB
r12
=
2
F1
r2
ke q A q B
2
F2 =
F1 r12
⎛ 13.7 mm ⎞
= 1.57 µ N
= 2.62 µ N ⎜
⎝ 17.7 mm ⎟⎠
r22
Then FB on A ,2 = 1.57 µ N to the left .
P23.5
)
(a)
( 8.99 × 109 N ⋅ m 2 C2 (1.60 × 10−19 C
kqq
Fe = e 12 2 =
2
r
( 3.80 × 10−10 m
(b)
Fg =
Gm1 m2 ( 6.67 × 10
=
r2
−11
)
)
)
2
= 1.59 × 10 −9 N
N ⋅ m 2 C2 (1.67 × 10 −27 kg
( 3.80 × 10
−10
m
)
2
)
( repulsion )
2
= 1.29 × 10 −45 N
The electric force is larger by 1.24 × 10 36 times .
(c)
P23.6
If ke
mm
q1q2
= G 1 2 2 with q1 = q2 = q and m1 = m2 = m , then
r
r2
q
=
m
G
6.67 × 10 −11 N ⋅ m 2 kg2
=
= 8.61 × 10 −11 C kg
ke
8.99 × 10 9 N ⋅ m 2 C2
We find the equal-magnitude charges on both spheres:
F = ke
q1q2
q2
=
k
e 2
r2
r
so
q=r
1.00 × 10 4 N
F
= 1.05 × 10 −3 C
= (1.00 m )
ke
8.99 × 10 9 N ⋅ m 2 C2
The number of electron transferred is then
N xfer =
1.05 × 10 −3 C
= 6.59 × 1015 electrons
1.60 × 10 −19 C e−
The whole number of electrons in each sphere is
⎛
10.0 g
⎞
N tot = ⎜
(6.02 × 10 23 atoms mol ) ( 47 e− atom ) = 2.62 × 10 24 e−
⎝ 107.87 g mol ⎟⎠
The fraction transferred is then
f=
N xfer ⎛ 6.59 × 1015 ⎞
= 2.51 × 10 −9 = 2.51 charges in every billion
=
N tot ⎜⎝ 2.62 × 10 24 ⎟⎠
5
6
P23.7
Chapter 23
F1 = ke
2
2
−6
−6
9
q1q2 (8.99 × 10 N ⋅ m C ) ( 7.00 × 10 C ) ( 2.00 × 10 C )
=
= 0.503 N
r2
( 0.500 m )2
F2 = ke
2
2
−6
−6
9
q1q2 (8.99 × 10 N ⋅ m C ) ( 7.00 × 10 C ) ( 4.00 × 10 C )
=
= 1.01 N
2
2
r
( 0.500 m )
Fx = 0.503 cos 60.0° + 1.01 cos 60.0° = 0.755 N
Fy = 0.503 sin 60.0° − 1.01 sin 60.0° = −0.436 N
F = ( 0.755 N ) ˆi − ( 0.436 N ) ˆj = 0.872 N at an angle of 3300°
FIG. P23.7
P23.8
Let the third bead have charge Q and be located distance x from the left end of the rod. This bead
will experience a net force given by
k ( 3q ) Q
ˆi + ke ( q ) Q − ˆi
F= e 2
x
( d − x )2
( )
1 , or d − x = x .
The net force will be zero if 3 =
2
3
x
( d − x )2
This gives an equilibrium position of the third bead of x = 0.634 d .
The equilibrium is stable if the third bead has positive chargee .
P23.9
(a)
The force is one of attraction . The distance r in Coulomb’s law is the distance between
centers. The magnitude of the force is
F=
(b)
The net charge of −6.00 × 10 −9 C will be equally split between the two spheres, or
−3.00 × 10 −9 C on each. The force is one of repulsion , and its magnitude is
F=
P23.10
12.0 × 10 −9 C ) (18.0 × 10 −9 C )
ke q1q2
9
2
2 (
=
8
.
99
×
10
N
⋅
m
C
= 2.16 × 10 −5 N
(
)
r2
( 0.300 m )2
3.00 × 10 −9 C ) ( 3.00 × 10 −9 C )
ke q1q2
9
2
2 (
=
8
.
99
×
10
N
⋅
m
C
= 8.99 × 10 −7 N
(
)
r2
( 0.300 m )2
The top charge exerts a force on the negative charge
ke qQ
( d /2)2 + x 2
which is directed upward and to
⎛ d⎞
the left, at an angle of tan −1 ⎜ ⎟ to the x-axis. The two positive charges together exert force
⎝ 2x ⎠
⎛ 2 k qQ ⎞ ⎛
( − x ) î
e
⎜ 2
⎜ 2
2 ⎟
⎝ ( d 4 + x ⎠ ⎜⎝ ( d 4 + x 2
)
(a)
)
−2 ke qQ a≈
x
md 3 8
The acceleration is equal to a negative constant times the excursion from equilibrium, as in
16 ke qQ
a = −ω 2 x , so we have Simple Harmonic Motion with ω 2 =
.
md 3
T=
(b)
⎞
d
,
1 2 ⎟ = ma or for x <<
⎟⎠
2
π
2π
=
ω
2
md 3
, where m is the mass of the object with charge −Q.
ke qQ
vmax = ω A = 4 a
ke qQ
md 3
Electric Fields
P23.11
(a)
k e2
F = e 2 = ( 8.99 × 10 9 N ⋅ m 2 C2
r
)
(1.60 × 10
−19
( 0.529 × 10
C
−10
)
m
7
2
)
2
= 8.22 × 10 −8 N toward the other
particle
We have F =
(b)
mv 2
from which
r
v=
Section 23.4
P23.12
8.22 × 10 −8 N ( 0.529 × 10 −10 m
Fr
=
m
9.11 × 10
−31
kg
)=
2.19 × 10 6 m s
The Electric Field
The point is designated in the sketch. The magnitudes of the
electric fields, E1 (due to the −2.50 × 10 −6 C charge) and E2
(due to the 6.00 × 10 −6 C charge), are
)
)
(1)
)
)
(2)
9
2
2
−6
ke q ( 8.99 × 10 N ⋅ m C ( 2.50 × 10 C
=
r2
d2
E1 =
E2 =
9
2
2
−6
ke q ( 8.99 × 10 N ⋅ m C ( 6.00 × 10 C
=
2
r2
( d + 1.00 m )
FIG. P23.12
Equate the right sides of (1) and (2)
to get
( d + 1.00 m )2 = 2.40d 2
or
d + 1.00 m = ±1.55 d
which yields
d = 1.82 m
or
d = −0.392 m
The negative value for d is unsatisfactory because that locates a point between the charges where
both fields are in the same direction.
Thus,
d = 1.82 m to the left of the − 2.50 µ C charge
8
P23.13
P23.14
Chapter 23
For equilibrium,
Fe = − Fg
or
qE = − mg − ˆj
Thus,
mg
ˆj
E=
q
( )
)
)
(a)
mg
9.11 × 10 −31 kg ( 9.80 m s 2
ˆj = (
ˆj = − ( 5.58 × 10 −11 N C ˆj
E=
q
( −1.60 × 10−19 C
(b)
mg
1.67 × 10 −27 kg ( 9.80 m s 2
ˆj = (
ˆj =
E=
q
(1.60 × 10−19 C
)
)
)
)
)
)
( 8.99 × 109 N ⋅ m 2 C2 (1.60 × 10−19 C
qq
F = ke 1 2 2 =
2
r
⎡⎣ 2 ( 6.37 × 10 6 m ⎤⎦
)
(1.02 × 10
−7
)
N C ˆj
) ( 6.02 × 10 )
2
23 2
= 514 kN
keQ
to the right.
+Q x = 0
a2
Both charges are on the x axis, so the total field cannot have a
vertical comonent, but it can be either to the right or to the left.
FIG. P23.15
If the total field at the origin is to the right, then q must be negative:
*P23.15 The first charge creates at the origin field
( )
keQ ˆ ke q
ˆ 2 keQ ˆi
i+
2 −i =
2
a
a2
( 3a )
q = −9Q
In the alternative, if the total field at the origin is to the left,
( )
( )
keQ ˆ ke q ˆ 2 keQ ˆ
i + 2 −i = 2 −i
a2
9a
a
P23.16
(a)
E=
)
)
9
−6
ke q ( 8.99 × 10 ( 2.00 × 10
=
= 14 400 N C
r2
(1.12 )2
Ex = 0
and
so
P23.17
q = +27Q
Ey = 2 (14 400 ) sin 26.6° = 1.29 × 10 4 N C
FIG. P23.16
E = 1.29 × 10 4 ˆj N C
(b)
F = qE = ( −3.00 × 10 −6 1.29 × 10 4 ˆj = −3.86 × 10 −2 ĵj N
(a)
kq
k ( 3q )
k ( 4q)
k ( 2q )
kq
kq
E = e 2 1 rˆ1 + e 2 2 rˆ2 + e 2 3 rˆ3 = e 2 ˆi + e 2 ˆi cos 45.0° + ˆj sin 45.0° + e 2 ĵ
2a
r1
r2
r3
a
a
)(
)
(
kq
kq
kq
E = 3.06 e2 ˆi + 5.06 e2 ˆj = 5.91 e2 at 58.88°
a
a
a
(b)
k q2
F = qE = 5.91 e 2 at 58.8°
a
)
q
x
Electric Fields
P23.18
The electric field at any point x has the x-component
E=−
ke q ( 4 ax )
ke q
ke q
2 +
2 = −
2
( x − a ) ( x − ( − a ))
( x 2 − a2
)
When x is much, much greater than a, we find E ≈ −
P23.19
4 a ( ke q )
.
x3
kQ n
One of the charges creates at P a field E = 2e
at an angle q
R + x2
to the x-axis as shown.
(a)
When all the charges produce field, for n > 1, the components
perpendicular to the x-axis add to zero.
The total field is
(b)
nke (Q n ) ˆi
cos θ =
R2 + x 2
(R
keQxˆi
2
+ x2
)
32
.
FIG. P23.19
A circle of charge corresponds to letting n grow beyond all bounds, but the result does
not depend on n. Smearing the charge around the circle does not change its amount or its
distance from the field point, so it does not change the field .
Section 23.5
P23.20
Electric Field of a Continuous Charge Distribution
E=∫
ke dq
, where dq = λ0 dx
x2
∞
dx
⎛ 1⎞
E = ke λ 0 ∫ 2 = ke λ 0 ⎜ − ⎟
⎝ x⎠
x
x0
P23.21
E=
∞
=
x0
ke λ 0
x0
The direction is − î or left for λ0 > 0.
)
k (Q ) ( 8.99 × 109 ( 22.0 × 10−6
ke λ keQ
= e
=
=
d ( + d ) d ( + d ) d ( + d ) ( 0.290 ) ( 0.140 + 0.290 )
E = 1.59 × 10 6 N C , directed toward the rod .
P23.22
E=
(x
keQx
2
+ a2
)
FIG. P23.21
32
⎡
1
For a maximum, dE = Qk ⎢
e
dx
⎢ ( x 2 + a2
⎣
x 2 + a 2 − 3x 2 = 0 or x =
)
32
−
3x 2
(x
2
+a
)
2 52
a
2
Substituting into the expression for E gives
E=
keQa
2( a
3
2
)
2 32
=
)
keQ
2 keQ
Q
=
=
2
3 2
32 a
3 3a
6 3π 0 a 2
⎤
⎥=0
⎥
⎦
9
10
P23.23
Chapter 23
E=
(x
ke xQ
2
+ a2
( 8.99 × 10 ) ( 75.0 × 10 ) x = 6.74 × 10 x
( x + 0.1100 )
( x + 0.010 0 )
−6
9
)
32
=
5
2 32
2
2
32
We choose the x axis along
the axis of the ring.
P23.24
(a)
At x = 0.010 0 m ,
E = 6.64 × 10 6 ˆi N C = 6.64 ˆi MN C
(b)
At x = 0.050 0 m ,
E = 2.41 × 10 7 ˆi N C = 24.1ˆi MN C
(c)
At x = 0.300 m,
E = 6.40 × 10 6 ˆi N C = 6.40 ˆi MN C
(d)
At x = 1.00 m ,
E = 6.64 × 10 5 ˆi N C = 0.664 ˆi MN C
⎛
⎞
x
E = 2π keσ ⎜ 1 − 2
2 ⎟
⎝
x +R ⎠
⎛
⎞
⎛
⎞
x
x
⎟ = 4.46 × 10 8 ⎜ 1 −
E = 2π ( 8.99 × 10 9 ( 7.90 × 10 −3 ⎜ 1 −
⎟
2
2
⎜⎝
⎝
x + 0.123 ⎠
x 2 + ( 0.350 ) ⎟⎠
)
P23.25
)
(a)
At x = 0.050 0 m ,
E = 3.83 × 10 8 N C = 383 MN C
(b)
At x = 0.100 m ,
E = 3.24 × 10 8 N C = 324 MN C
(c)
At x = 0.500 m,
E = 8.07 × 10 7 N C = 80.7 MN C
(d)
At x = 2.000 m,
E = 6.68 × 10 8 N C = 6.68 MN C
(a)
⎛
⎞
x
From the Example in the chapter text, E = 2π keσ ⎜ 1 −
2
2 ⎟
⎝
x +R ⎠
σ=
Q
= 1.84 × 10 −3 C m 2
π R2
)
E = (1.04 × 10 8 N C ( 0.900 ) = 9.36 × 10 7 N C = 93.6 MN C
approximation: E = 2π keσ = 104 MN C ( about 11% high )
(b)
⎛
⎞
30.0 cm
E = (1.04 × 10 8 N C ⎜ 1 −
= (1.04 10 8 N C ( 0.004 96 ) = 0.516 MN C
⎟
2
2
⎝
30.0 + 3.00 cm ⎠
)
approximation: E = ke
)
Q
5.20 × 10 −6
= ( 8.99 × 10 9
= 0.519 MN C ( about 0.6% high )
2
r
( 0.30 )2
)
Electric Fields
P23.26
The electric field at a distance x is
⎡
⎤
x
Ex = 2π keσ ⎢1 −
⎥
2
2
x +R ⎦
⎣
This is equivalent to
⎡
⎤
1
⎥
Ex = 2π keσ ⎢1 −
2
2
⎢⎣
1 + R x ⎥⎦
For large x,
R2
<< 1 and
x2
1+
R2
R2
≈ 1+ 2
2
x
2x
⎛
⎞
1
Ex = 2π keσ ⎜ 1 −
⎟
⎜⎝ ⎡1 + R 2 ( 2 x 2 ⎤ ⎟⎠
⎣
⎦
so
)
(1 + R ( 2 x ) − 1)
2
= 2π keσ
Substitute σ =
Q
,
π R2
But for x >> R,
P23.27
11
Ex =
1
1
≈ , so
x + R2 2 x 2
)
⎡⎣1 + R 2 ( 2 x 2 ⎤⎦
keQ (1 x 2
)
⎛ 2 R2 ⎞
=
k
Q
e ⎜x +
2 ⎟⎠
⎝
⎡⎣1 + R 2 ( 2 x 2 ⎤⎦
Ex ≈
2
2
)
keQ
for a disk at large distances
x2
Due to symmetry
Ey = ∫ dEy = 0, and Ex = ∫ dE sin θ = ke ∫
where
dq = λ ds = λ rdθ ,
so that
Ex =
dq sin θ
r2
π
π
2k λ
ke λ
kλ
sin θ dθ = e ( − cos θ ) 0 = e
r ∫0
r
r
q
L
and r =
L
π
where
λ=
Thus,
Ex =
Solving,
Ex = −2.16 × 10 7 N C
2
2
−6
9
2 ke qπ 2 (8.99 × 10 N m C ) ( −7.50 × 10 C ) π
=
2
L2
( 0.140 m )
Since the rod has a negative charge, E = −2.16 × 10 7 ˆi N C = −21.6 ˆi MN C .
(
)
FIG. P23.27
12
P23.28
Chapter 23
(a)
We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has
Qdx
and produces, at the chosen point, a field
h
ke x
Qdx ˆ
dE =
i
2
2 32
h
(x + R
charge
)
The total field is
E=
∫
d +h
dE =
all charge
keQxdx
∫ h(x
d
)
+R
−1 2 d + h
k Qˆi ( x 2 + R 2
E= e
2h
(− 1 2)
(b)
2
x=d
ˆ d +h
ˆi = keQi ( x 2 + R 2
2 h x∫= d
)
2 32
⎡
keQî ⎢
1
=
2
⎢
h ( d + R2
⎢⎣
So,
E=
)
12
−
2 xdx
(( d + h )
1
2
⎤
⎥
12 ⎥
2
+R
⎥⎦
)
d + h 2 keQdx ⎛
x
dE = ∫
⎜1 − 2
2
2
⎜
R
h
all charge
x=d
⎝ (x + R
∫
)
−1 2
2 k Qˆi
2
E = e2 ⎡ d + h − d − ( d + h ) + R 2
R h ⎣⎢
)
12
)
12
⎞
⎟ î
⎟⎠
⎡
⎤ 2 k Qî d + h 1 ( x 2 + R 2
2 xdx ⎥ = e2 ⎢ x d −
Rh ⎢
2
12
⎦
⎣⎢
) + (d
2
+ R2
− ( d + h ) + R2
)
⎤
⎦⎥
(
2 keQî ⎡
h + ( d 2 + R2
E=
R 2 h ⎣⎢
Qdx
, and chargeh
⎞
⎟ ˆi
⎟⎠
d +h
2 k Qˆi ⎡ d + h
1
E = e2 ⎢ ∫ dx − ∫ ( x 2 + R 2
Rh ⎣d
2 x=d
(a)
12
−3 2
Think of the cylinder as a stack of disks, each with thickness dx, charge
Qdx
per-area σ =
. One disk produces a field
π R2 h
2π keQdx ⎛
x
dE =
⎜1 −
π R 2 h ⎜⎝ ( x 2 + R 2
P23.29
)
)
(
12
2
12
)
12
1 2 d +h
)
d
⎤
⎥
⎥
⎥⎦
⎤
⎦⎥
The electric field at point P due to each element of length dx is
k dq
dE = 2 e 2 and is directed along the line joining the element to
x +y
point P. By symmetry,
Ex = ∫ dEx = 0
and since
dq = λ dx ,
E = Ey = ∫ dEy = ∫ dE cos θ
where
cos θ =
y
x + y2
2
FIG. P23.29
2
Therefore,
E = 2 ke λ y ∫
0
(b)
For a bar of infinite length,
dx
(x
2
+y
)
2 32
θ 0 = 90°
=
2 ke λ sin θ 0
y
and
Ey =
2 ke λ
y
Electric Fields
P23.30
(a)
The whole surface area of the cylinder is A = 2π r 2 + 2π rL = 2π r ( r + L ).
)
Q = σ A = (15.0 × 10 −9 C m 2 2π ( 0.025 0 m ) [ 0.025 0 m + 0.060 0 m ] = 2.00 × 10 −10 C
(b)
For the curved lateral surface only, A = 2π rL.
)
Q = σ A = (15.0 × 10 −9 C m 2 ⎡⎣ 2π ( 0.025 0 m ) ( 0.060 0 m ) ⎤⎦ = 1.41 × 10 −10 C
P23.31
)
(c)
2
Q = ρV = ρπ r 2 L = ( 500 × 10 −9 C m 3 ⎡π ( 0.025 0 m ) ( 0.060 0 m ) ⎤ = 5.89 × 10 −11 C
⎣
⎦
(a)
−4
3
Every object has the same volume, V = 8 ( 0.030 0 m ) = 2.16 × 10 m .
3
)
)
For each, Q = ρV = ( 400 × 10 −9 C m 3 ( 2.16 × 10 −4 m 3 = 8.64 × 10 −11 C
(b)
We must count the 9.00 cm2 squares painted with charge:
(i)
6 × 4 = 24 squares
)
)
)
)
)
)
)
)
Q = σ A = (15.0 × 10 −9 C m 2 24.0 ( 9.00 × 10 −4 m 2 = 3.24 × 10 −10 C
(ii)
34 squares exposed
Q = σ A = (15.0 × 10 −9 C m 2 24.0 ( 9.00 × 10 −4 m 2 = 3.24 × 10 −10 C
(iii)
34 squares
Q = σ A = (15.0 × 10 −9 C m 2 34.0 ( 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C
(iv)
32 squares
Q = σ A = (15.0 × 10 −9 C m 2 32.0 ( 9.00 × 10 −4 m 2 = 4.32 × 10 −10 C
(c)
(i)
total edge length: = 24 × ( 0.030 0 m )
)
Q = λ = ( 80.0 × 10 −12 C m 24 × ( 0.030 0 m ) = 5.76 × 10 −11 C
Section 23.6
(ii)
Q = λ = (80.0 × 10 −12 C m ) 44 × ( 0.030 0 m ) = 1.06 × 10 −10 C
(iii)
Q = λ = ( 80.0 × 10 −12 C m 64 × ( 0.030 0 m ) = 1.54 × 10 −10 C
(iv)
Q = λ = ( 80.0 × 10 −12 C m 40 × ( 0.030 0 m ) = 0.960 × 10 −10 C
)
)
Electric Field Lines
P23.32
P23.33
FIG. P23.32
FIG. P23.33
13
14
P23.34
Chapter 23
(a)
(b)
P23.35
(a)
q1 −6
1
=
= −
q2 18
3
q1 is negative, q2 is positive
The electric field has the general appearance shown. It is
zero at the center , where (by symmetry) one can see that
the three charges individually produce fields that cancel out.
In addition to the center of the triangle, the electric field lines in
the second figure to the right indicate three other points near the
middle of each leg of the triangle where E = 0, but they are
more difficult to find mathematically.
(b)
You may need to review vector addition in Chapter Three.
The electric field at point P can be found by adding the
electric field vectors
due to each of the two lower point
charges: E = E1 + E 2 .
q
The electric field from a point charge is E = ke 2 rˆ .
r
As shown in the solution figure at right,
q
E1 = ke 2 to the right and upward at 60°
a
q
E 2 = ke 2 to the left and upward at 60°
a
FIG. P23.35
q
q
E = E1 + E 2 = ke 2 ⎡ cos 60°ˆi + sin 60°ˆj + − cos 60°ˆi + sin 60°ˆj ⎤ = ke 2 ⎡ 2 sin 60°ˆj ⎤
⎦
⎦
a ⎣
a ⎣
(
= 1.73ke
Section 23.7
P23.36
)
(
)
q
ĵ
a2
Motion of a Charged Particle in a Uniform Electric Field
−19
5
qE (1.602 × 10 ) ( 6.00 × 10 )
=
= 5.76 × 1013 m s so a = −5.76 × 1013 ˆi m s 2
m
(1.67 × 10 −27 )
(a)
a =
(b)
v 2f = v i2 + 2 a x f − xi
(
)
0 = vi2 + 2 ( −5.76 × 1013 ) ( 0.070 0 )
(c)
) (
v i = 2.84 × 10 6 ˆi m s
v f = vi + at
0 = 2.84 × 10 6 + ( −5.76 × 1013 ) t
t = 4.93 × 10 −8 s
Electric Fields
P23.37
P23.38
15
qE 1.602 × 10 −19 ( 640 )
=
= 6.14 × 1010 m s 2
m
1.67 × 10 −27
(a)
a=
(b)
v f = vi + at
(c)
x f − xi =
(d)
K=
1.20 × 10 6 = ( 6.14 × 1010 ) t
)
(
1
vi + v f t
2
xf =
t = 1.95 × 10 −5 s
1
(1.20 × 106 ) (1.95 × 10 −5 ) = 11.7 m
2
2
1
1
m v 2 = (1.67 × 10 −27 kg ) (1.20 × 10 6 m s ) = 1.20 × 10 −15 J
2
2
The particle feels a constant force:
F = qE = (1 × 10 −6 C ) ( 2 000 N C ) − ˆj = 2 × 10 −3 N − ˆj
and moves with acceleration:
a=
∑ F = ( 2 × 10
−3
( )
kg ⋅ m s ) ( − ˆj)
= (1 × 10
2
13
2 × 10 −16 kg
m
( )
( )
m s 2 ) − ĵ
Note that the gravitational force is on the order of a trillion times smaller than the electrical force
exerted on the particle. The particle’s x-component of velocity is constant at
(1.00 × 10
5
m s ) cos 37° = 7.99 × 10 4 m s . Thus it moves in a parabola opening downward. The
maximum height it attains above the bottom plate is described by
)
(
)
)(
0 = ( 6.02 × 10 4 m s − ( 2 × 1013 m s 2 y f − 0
vyf2 = vyi2 + 2ay y f − yi :
2
)
y f = 1.81 × 10 −4 m
Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetric
parabola and strikes the bottom plate after a time given by
y f = yi + vyi t +
1 2
ay t
2
0 = 0 + ( 6.02 × 10 4 m s ) t +
1
( −1 × 1013 m s2 ) t 2
2
since t > 0,
t = 1.20 × 10 −8 s
The particle’s range is
x f = xi + vx t = 0 + ( 7.99 × 10 4 m s ) (1.20 × 10 −8 s ) = 9.61 × 10 −4 m.
In sum,
The particle strikes the negative plate after moving in a parabola with a height of 0.181 mm
and a width of 0.961 mm.
P23.39
The required electric field will be in the direction of motion .
Work done = ∆K
so,
1
− Fd = − m vi2 (since the final velocity = 0)
2
which becomes
eEd = K
and
E=
K
ed
16
P23.40
Chapter 23
vi = 9.55 × 10 3 m s
(a)
ay =
−19
eE (1.60 × 10 ) ( 720 )
=
= 6.90 × 1010 m s 2
m
(1.67 × 10 −27 )
From the large magnitude of this vertical acceleration,
we can note that the gravitational force on the particle is
negligible by comparison to the electrical force.
vi2 sin 2θ
= 1.27 × 10 −3 m so that
ay
R=
(9.55 × 10 )
3 2
sin 2θ
6.90 × 10
10
sin 2θ = 0.961
P23.41
= 1.27 × 10 −3
90.0° − θ = 53.1°
θ = 36.9°
(b)
t=
R
R
=
vix vi cos θ
(a)
t=
0.050 0
x
=
= 1.11 × 10 −7 s = 111 ns
vx 4.50 × 10 5
(b)
ay =
If θ = 36.9°, t = 167 ns .
)
If θ = 53.1° , t = 221 ns .
)
−19
3
qE (1.602 × 10 ( 9.60 × 10
=
= 9.21 × 1011 m s 2
m
(1.67 × 10−27
1
y f − yi = vyi t + ay t 2 :
2
(c)
FIG. P23.40
vx = 4.50 × 10 5 m s
)
yf =
2
1
9.21 × 1011 ) (1.11 × 10 −7 ) = 5.68 × 10 −3 m = 5.68 mm
(
2
vyf = vyi + ay t = ( 9.21 × 1011 ) (1.11 × 10 −7 ) = 1.02 × 10 5 m s
Additional Problems
P23.42
The two given charges exert equal-size forces of attraction
on each other. If a third charge, positive or negative, were
placed between them they could not be in equilibrium. If the
third charge were at a point x > 15 cm, it would exert a
stronger force on the 45 m C than on the −12 m C, and could
not produce equilibrium for both. Thus the third charge must
be at x = −d < 0. Its equilibrium requires
ke q (12 µ C ) ke q ( 45 µ C )
=
d2
(15 cm + d )2
⎛ 15 cm + d ⎞ = 45 = 3.75
⎝
⎠
12
d
15 cm + d = 1.94 d
d = 16.0 cm
d
q
x=0
–
–12 m C
15 cm
x
+
45 m C
FIG. P23.42
2
The third charge is at x = −16.0 cm . The equilibrium of the −12 m C requires
ke q (12 µ C ) ke ( 45 µ C ) 12 µ C
=
(16.0 cm )2
(15 cm )2
q = 51.3 µ C
All six individual forces are now equal in magnitude, so we have equilibrium as required, and this
is the only solution.
Electric Fields
P23.43
The proton moves with acceleration a p =
while the e− has acceleration
(a)
Thus,
(a)
(1.60 × 10
−19
C ) ( 640 N C )
9.110 × 10 −31 kg
d=
= 1.12 × 1014 m s 2 = 1 836a p
1 2
a p t ), knowing:
2
1 2 1 2
1
a p t + ae t = 1 837 ⎛ a p t 2 ⎞
⎝
⎠
2
2
2
1 2 4.00 cm
apt =
= 21.8 µ m
2
1 837
The distance from the positive plate to where the meeting occurs equals the distance the
1
sodium ion travels (i.e., dNa = aNa t 2 ). This is found from:
2
4.00 cm =
P23.44
)
−19
qE (1.60 × 10 C ( 640 N C )
=
= 6.13 × 1010 m s 2
1.673 × 10 −27 kg
m
We want to find the distance traveled by the proton (i.e., d =
4.00 cm =
(b)
ae =
17
1
1
aNa t 2 + aCl t 2 :
2
2
1
eE ⎞ 2 1 ⎛ eE ⎞ 2
4.00 cm = ⎛
t +
t
⎝
2 22.99 u ⎠
2 ⎝ 35.45 u ⎠
This may be written as
4.00 cm =
so
dNa =
1
1
1
aNa t 2 + ( 0.649aNa ) t 2 = 1.65 ⎛ aNa t 2 ⎞
⎝2
⎠
2
2
1
4.00 cm
= 2.43 cm
aNa t 2 =
2
1.65
The field, E1, due to the 4.00 × 10 −9 C charge is in the
−x direction.
(8.99 × 109 N ⋅ m 2 C2 ) ( −4.00 × 10 −9 C) ˆi
kq
E1 = e2 rˆ =
r
( 2.50 m )2
= −5.75ˆi N C
FIG. P23.44 (a)
Likewise, E2 and E3 , due to the 5.00 × 10 −9 C charge and the 3.00 × 10 −9 C charge, are
8.99 × 10 9 N ⋅ m 2 C2 ) ( 5.00 × 10 −9 C ) ˆ
(
ke q
ˆ
E2 = 2 r =
i = 11.2 N C ˆi
r
( 2.00 m )2
(8.99 × 109 N ⋅ m 2 C2 ) (3.00 × 10 −9 C) ˆi = 18.7 N C ˆi
E3 =
(1.20 m )2
E R = E1 + E 2 + E 3 = 24.2 N C in +x direction.
continued on next page
18
Chapter 23
(b)
kq
E1 = e2 rˆ = ( −8.46 N C ) 0.243ˆi + 0.970 ˆj
r
(
)
kq
E 2 = e2 rˆ = (11.2 N C ) + ˆj
r
ke q
E 3 = 2 rˆ = ( 5.81 N C ) −0.371ˆi + 0.928ˆj
r
Ex = E1x + E3 x = −4.21ˆi N C
Ey = E1y + E2 y + E3 y = 8.43 ĵ N C
( )
(
(a)
FIG. P23.44 (b)
θ = 63.4 º above − x axis
ER = 9.42 N C
P23.45
)
Let us sum force components to find
∑F
x
= qEx − T sin θ = 0 , and
∑F
y
= qEy + T cos θ − mg = 0
Combining these two equations, we get
q=
(
(1.00 × 10 −3 ) (9.80 ) = 1.09 × 10 −8 C
mg
=
( 3.00 cot 37.0° + 5.00 ) × 10 5
Ex cot θ + Ey
)
= 10.9 nC
(b)
From the two equations for
T=
P23.46
∑F
x
and
∑F
y
FIG. P23.45
we also find
qEx
= 5.44 × 10 −3 N = 5.44 mN
sin 37.0°
This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ.
The unknowns are q and T.
The approach to this problem should be the same as for the last problem, but without numbers to
substitute for the variables. Likewise, we can use the free body diagram given in the solution to
problem 45.
Again, Newton’s second law:
∑F
= −T sin θ + qA = 0
(1)
and
∑F
= +T cos θ + qB − mg = 0
(2)
(a)
(b)
x
y
Substituting T =
qA
into Eq. (2),
sin θ
qA cos θ
+ qB = mg
sin θ
Isolating q on the left,
q=
Substituting this value into Eq. (1),
T=
mg
( A cot θ + B )
mgA
( A cos θ + B sin θ )
If we had solved this general problem first, we would only need to substitute the appropriate
values in the equations for q and T to find the numerical results needed for problem 45. If you
find this problem more difficult than problem 45, the little list at the first step is useful. It shows
what symbols to think of as known data, and what to consider unknown. The list is a guide for
deciding what to solve for in the analysis step, and for recognizing when we have an answer.
Electric Fields
P23.47
F=
ke q1q2
:
r2
15.0
60.0
θ = 14.0°
tan θ =
(8.99 × 10 ) (10.0 × 10 )
F =
= 40.0 N
(8.99 × 10 ) (10.0 × 10 )
= 2.50 N
(8.99 × 10 ) (10.0 × 10 )
=
= 2.35 N
−6 2
9
( 0.150 )2
1
−6 2
9
F3 =
( 0.600 )2
−6 2
9
F2
FIG. P23.47
( 0.619 )2
Fx = − F3 − F2 cos 14.0° = −2.50 − 2.35 cos 14.0° = −4.78 N
Fy = − F1 − F2 sin 14.0° = −40.0 − 2.35 sin 14.0° = −40.6 N
Fnet = Fx2 + Fy2 = ( 4.78 ) + ( 40.6 ) = 40.9 N
2
tan φ =
Fy
Fx
=
2
−40.6
−4.78
φ = 263°
P23.48
From Figure (a) we have
d cos 30.0° = 15.0 cm
d=
or
From Figure (b) we have
15.0 cm
cos 30.0°
d ⎞
θ = sin −1 ⎛
⎝ 50.0 cm ⎠
⎛
⎞
15.0 cm
= 20.3°
θ = sin −1 ⎜
⎝ 50.0 cm ( cos 30.0° ) ⎟⎠
Fq
mg
= tan θ
Fq = mg tan 20.3°
or
From Figure (c) we have
Figure (a)
(1)
Fq = 2 F cos 30.0°
⎤
⎡
ke q 2
Fq = 2 ⎢
2 ⎥ cos 30.0°
⎣ ( 0.300 m ) ⎦
(2)
Figure (b)
Combining equations (1) and (2),
⎡
⎤
ke q 2
2⎢
2 ⎥ cos 30.0° = mg tan 20.3°
)
.
m
0
300
(
⎣
⎦
q2 =
mg ( 0.300 m ) tan 20.3°
2 ke cos 30.0°
q2 =
( 2.00 × 10 kg) (9.80 m s ) ( 0.300 m ) tan 20.3°
2 (8.99 × 10 N ⋅ m C ) cos 30.0°
2
−3
2
2
9
2
2
q = 4.20 × 10 −14 C2 = 2.05 × 10 −7 C = 0.205 µ C
Figure (c)
FIG. P23.48
19
20
P23.49
Chapter 23
Charge
F=
Q
resides on each of the blocks, which repel as point charges:
2
ke ( Q 2 ) ( Q 2 )
= k ( L − Li )
L2
Solving for Q,
P23.50
k ( L − Li )
ke
Q = 2L
If we place one more charge q at the 29th vertex, the total force on the central charge will add up
k qQ
to zero: F28 charges + e 2 away from vertex 29 = 0
a
P23.51
k qQ
F28 charges = e 2 toward vertex 29
a
According to the result of an Example in the chapter text, the lefthand rod creates this field at a distance d from its right-hand end:
E=
keQ
d ( 2a + d )
dF =
FIG. P23.51
keQQ
dx
2a d ( d + 2a )
k Q2
1
2a + x ⎞
dx
= e ⎛−
ln
∫
⎝
)
+
2
a
2
a
2
a
x ⎠ b− 2 a
x
x
(
x =b− 2 a
b
b
F=
keQ 2
2a
F=
⎛ k Q2 ⎞ ⎛ b2 ⎞
2a + b
+ keQ 2 ⎛
b ⎞ keQ 2
b2
ln
− ln
+ ln
=
= ⎜ e 2 ⎟ ln ⎜ 2
2 ⎝
2
4a
b
b − 2a ⎠ 4 a
( b − 2a ) ( b + 2a ) ⎝ 4 a ⎠ ⎝ b − 4 a 2 ⎟⎠
*P23.52 We model the spheres as particles. They have different charges. They exert on each other forces
of equal magnitude. They have equal masses, so their strings make equal angles q with the
vertical. The distance r between them is described by sin q = (r/2)/40 cm,
so
r = 80 cm sin q
Let T represent the string tension. We have
ΣFx = 0:
keq1q2/r2 = T sinq
ΣFy = 0: mg = T cosq
Divide to eliminate T.
ke q1q2
r /2
= tan θ =
2
r mg
(40 cm)2 − r 2 / 4
Cleared of fractions,
ke q1q2 (80 cm)2 − r 2 = mgr 3
8.99 × 109(N ⋅ m2/C2) 300 × 10−9C (200 × 10−9C)
(0.8 m)2 − r 2 = 1901 r 6
We home in on a solution by trying values.
r
0
0.5
0.2
0.3
0.24
0.27
0.258
0.259
0.64 − r 2 − 1901 r 6
+0.64
−29.3
+0.48
−0.84
+0.22
−0.17
+0.013
−0.001
Thus the distance to three digits is 0.259 m .
(0.8 m)2 − r 2 = 2.4 × 10 −3 (9.8) N r 3
Electric Fields
*P23.53
Q = ∫ λd =
90.0 º
∫
−90.0 º
λ0 cos θ Rdθ = λ0 R sin θ −90.0 º = λ0 R [1 − ( −1)] = 2λ0 R
90.0 º
Q = 12.0 µ C = ( 2λ0 ) ( 0.600 ) m = 12.0 µ C
dFy =
so
λ0 = 10.0 µ C m
2
1 ⎛ ( 3.00 µ C ) ( λ0 cos θ Rdθ ) ⎞
1 ⎛ ( 3.00 µ C ) ( λ d ) ⎞
=
cos
θ
⎟
⎟⎠
R2
4π ∈0 ⎜⎝
R2
4π ∈0 ⎜⎝
⎠
90.0 º
Fy =
∫ (8.99 × 10
9
−90.0 º
N ⋅ m 2 C2 )
(3.00 × 10
−6
C ) (10.0 × 10 −6 C m )
( 0.600 m )
cos 2 θ dθ
1
8.99 ( 30.0 ) −3
(10 N ) ∫ ⎛⎝ 12 + 12 cos 2θ ⎞⎠ dθ
0.600
−π 2
1
1
Fy = ( 0.450 N ) ⎛ θ + sin 2θ ⎞
⎝2
⎠
4
π 2
−π 2
= 0.707 N downward
cos q
0
–1
π 2
Fy =
21
1
0
0°
360°
cos2 q
0°
360°
FIG. P23.53
Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0.
*P23.54 (a) The two charges create fields of equal magnitude, both with outward components along the
x axis and with upward and downward y components that add to zero. The net field is then
ke q x ˆ ke q x ˆ 2(8.99 × 10 9 N ⋅ m 2 )52 × 10 −9 C x ˆi
i+ 2 i=
r2 r
r r
C2 ((0.25 m)2 + x 2 )3/2
=
935 N ⋅ m 2 x ˆi
C(0.0625 m 2 + x 2 )3/ 2
(b) At x = 0.36 m,
E=
935 N ⋅ m 2 0.36 m ˆi
= 4.00 kN/C ˆi
C (0.0625 m 2 + (0.36 m)2 )3/2
(c) We solve 1 000 = 935 x (0.0625 + x2)−3/2 by tabulating values for the field function:
x
0
0.01
0.02
0.1
0.2
0.36
0.9
1
∞
935 x (0.0625 + x2)−3/2
0
597
1 185
4 789
5 698
4 000
1 032
854
0
We see that there are two points where E = 1 000. We home in on them to determine their
coordinates as (to three digits) x = 0.016 8 m and x = 0.916 m.
(d) The table in part (c) shows that the field is nowhere so large as 16 000 N/C.
(e) The field of a single charge in Question 7 takes on all values from zero to infinity, each
at just one point along the positive x axis. The vector sum of the field of two charges,
in this problem, is zero at the origin, rises to a maximum at 17.7 cm, and then decreases
asymptotically to zero. In the question and the problem, the fields at x = 36 cm happen
to take similar values. For large x the field of the two charges in this problem shows
just the same inverse proportionality to x2 as the field in the question, being larger
by the factor 2(52 nC) (57.7 nC) = 1.80 times.
/
22
P23.55
Chapter 23
(a)
From the 2Q charge we have
Fe − T2 sin θ 2 = 0 and mg − T2 cos θ 2 = 0
Combining these we find
Fe
T sin θ 2
= 2
= tan θ 2
mg T2 cos θ 2
From the Q charge we have
Fe = T1 sin θ1 = 0 and mg − T1 cos θ1 = 0
Combining these we find
Fe
T sin θ1
= 1
= tan θ1 or θ 2 = θ1
mg T1 cos θ1
Fe =
(b)
FIG. P23.55
ke 2QQ 2 keQ 2
=
r2
r2
If we assume q is small then
tan θ ≈
r 2
Substitute expressions for Fe and tan θ into either equation found in part (a) and solve for r.
2
⎛ 4 k Q2 ⎞
Fe
= tan θ , then 2 keQ ⎛ 1 ⎞ ≈ r and solving for r we find r ≈ ⎜ e
mg
⎝ mg ⎟⎠
r 2 ⎜⎝ mg ⎟⎠ 2 P23.56
.
The bowl exerts a normal force on each bead, directed along
the radius line or at 60.0° above the horizontal. Consider the
free-body diagram shown for the bead on the left side of the
bowl:
∑F
y
or
P23.57
13
= n sin 60.0° − mg = 0,
n=
mg
sin 60.0°
n
Also,
∑F
or
ke q 2
mg
mg
= n cos 60.0° =
=
R2
tan 60.0°
3
Thus,
⎛ mg ⎞
q = R⎜
⎟
⎝ ke 3 ⎠
(a)
x
= − Fe + n cos 60.0° = 0,
Fe
60.0°
mg
FIG. P23.56
12
qE ⎞
⎛
The total non-contact force on the cork ball is: F = qE + mg = m ⎜ g +
⎟,
⎝
m⎠
which is constant and directed downward. Therefore, it behaves like a simple pendulum in
the presence of a modified uniform gravitational field with a period given by:
T = 2π
L
0.500 m
= 2π
2
−6
g + qE m
9.80 m s + ⎡⎣( 2.00 × 10 C (1.00 × 10 5 N C 1.00 × 10 −3 kg ⎤⎦
)
)
= 0.307 s
(b)
Yes . Without gravity in part (a), we get T = 2π
T = 2π
( 2.00 × 10
−6
L
qE m
0.500 m
= 0.314 s (a 2.28% difference).
C ) (1.00 × 10 5 N C ) 1.00 × 10 −3 kg
Electric Fields
+ P
*P23.58 (a) At A the top charge makes the dominant contribution to the
field and the net field is downward. At B the total electric
field is zero. Between B and C the main change is weakening
of the downward electric field of the top charge, so the net field
at C is upward. At E the fields of the two bottom charges cancel
out and the total field is downward. At F the total field is
downward.
(b) The field is zero at B as it changes from downward at A to
upward at C. As a continuous function, the field must pass
through the value zero near D as it changes from upward
at C to downward at E and F.
A
B
C
D
+
y
(1.5 cm )
2
+y
2
+
E
+
F
FIG. P23.58
(c) Let y represent the distance from E up to the zero-field point.
The distance from P to E is (32 − 1.52)1/2 cm = 2.60 cm. Then
the requirement that the field be zero is
ke q
ke q
=
(2.60 cm − y )2 (1.5 cm)2 + y 2
23
ke qy
⎡⎣(1.5 cm)2 + y 2 ⎤⎦
3/ 2
ke q
2 ke qy
=
3/ 2
2
(2.60 cm − y )
⎡⎣(1.5 cm)2 + y 2 ⎤⎦
(1.52 + y2)3/2 − 2 y (2.60 − y)2 = 0
As a check on our algebra, we note that y = (1/3)2.60 cm = 0.866 cm should be a solution,
corresponding to point B. Substituting 0.866 gives 5.20 − 5.20 = 0 as it should. We home in
on the smaller answer:
(1.52 + y2)3/2 − 2y(2.60 − y)2
y
+3.375
+0.411
−0.124
+0.014
−0.000 6
0
0.3
0.4
0.37
0.373
To three digits the answer is 0.373 cm .
P23.59
(a)
There are 7 terms which contribute:
3 are s away (along sides)
3 are
1 is
2s away (face diagonals) and sin θ =
3s away (body diagonal) and sin φ =
1
= cos θ
2
1
3
FIG. P23.59
The component in each direction is the same by symmetry.
k q2
F = e2
s
(b)
)
2
1 ⎤ ˆ ˆ ˆ
ke q 2
⎡
1
=
+
+
i
+
j
+
k
(1.990 ) ˆi + ˆj + kˆ
⎢ 2 2 3 3⎥
s2
⎣
⎦
(
F = Fx2 + Fy2 + Fz2 = 3.29
(
ke q 2
away from the origiin
s2
)
24
P23.60
Chapter 23
(a)
Zero contribution from the same face due to symmetry, opposite
face contributes
2
⎛k q
⎞
4 ⎜ e2 sin φ ⎟
⎝ r
⎠
sin φ =
(b)
P23.61
where
s
r
2
⎛ s⎞ ⎛ s⎞
r = ⎜ ⎟ + ⎜ ⎟ + s 2 = 1.5 s = 1.22 s
⎝ 2⎠ ⎝ 2⎠
E=4
ke qs
4 ke q
kq
=
= 2.18 e2
3
2
r3
s
s
(1.22 )
FIG. P23.60
The direction is the k̂ direction.
The field on the axis of the ring is calculated in an Example in the chapter text as
E = Ex =
(x
ke xQ
2
+ a2
)
32
The force experienced by a charge −q placed along the axis of the ring is
⎡
x
F = − keQq ⎢
2
⎢ ( x + a2
⎣
)
32
⎤
k Qq
⎥ and when x << a, this becomes F = − ⎛ e 3 ⎞ x.
⎝
a ⎠
⎥
⎦
This expression for the force is in the form of Hooke’s law, with an effective spring
k Qq
constant of k = e 3 .
a
Since ω = 2π f =
P23.62
dE =
=
E=
k
1
, we have f =
m
2π
ke dq
2
2
x + ( 0.150 m )
(
keQq
.
ma 3
⎛ − xˆi + 0.150 mˆj ⎞
⎜
⎟
⎜⎝ x 2 + ( 0.150 m )2 ⎟⎠
)
ke λ − xˆi + 0.150 mˆj dx
⎡ x + ( 0.150 m ) ⎤
⎣
⎦
2
2
32
)
(
FIG. P23.62
0.400 m
− xˆi + 0.150 mˆj dx
dE = ke λ ∫
∫
2 32
2
all charge
x = 0 ⎡ x + ( 0.150 m ) ⎤
⎣
⎦
0.400 m
0.400 m
⎡
⎤
ˆjx
ˆi
0
150
m
.
(
+
)
⎢
⎥
+
E = ke λ
2
2
2
⎢ 2
⎥
2
+
x
x
0
150
m
0
.
150
m
+
0
.
150
m
.
(
(
(
)
)
)
⎢⎣
⎥⎦
0
0
E = ( 8.99 × 10 9 N ⋅ m 2 C2 ( 35.0 × 10 −9 C m ⎡⎣ ˆi ( 2.34 − 6.67 ) m −1 + ˆj ( 6.24 − 0 ) m −1 ⎤⎦
E = −1.36 ˆi + 1.96 ˆj × 10 3 N C = −1.36 ˆi + 1.96 ˆj kN C
)
(
)
)
(
)
Electric Fields
P23.63
The electrostatic forces exerted on the two charges result in a net torque
τ = −2 Fa sin θ = −2 Eqa sin θ .
For small θ , sin θ ≈ θ and using p = 2qa, we have
τ = −Epθ
The torque produces an angular acceleration given by
τ = Iα = I
Combining these two expressions for torque, we have
d 2θ ⎛ Ep ⎞
+⎜ ⎟θ = 0
dt 2 ⎝ I ⎠
d 2θ
dt 2
FIG. P23.63
d 2θ
= −ω 2θ which is the standard
dt 2
Ep
equation characterizing simple harmonic motion, with ω 2 =
I
This equation can be written in the form
Then the frequency of oscillation is f = w /2p, or
f =
1
2π
pE
1
=
I
2π
2 qaE
I
P23.64
− k qˆi
1 1
kq
kq
kq
kq
π 2 ke q
E = ∑ e2 rˆ = e2 − ˆi + e 2 − ˆi + e 2 − ˆi + … = e2 ⎛ 1 + 2 + 2 + …⎞ = −
î
⎠
a ⎝ 2 3
6a 2
r
a
( 2a )
( 3a )
P23.65
E = ∫ dE =
( )
( )
( )
( )
∞
⎡ ke λ0 x0 dx − ˆi ⎤
⎛ 1
⎥ = − ke λ0 x0 ˆi ∫ x −3 dx = − ke λ0 x0 ˆi ⎜ − 2
3
∫x ⎢⎢
x
⎥
⎝ 2x
x0
0 ⎣
⎦
∞
∞
⎞
ke λ 0
⎟ = 2 x − î
x0 ⎠
0
( )
ANSWERS TO EVEN PROBLEMS
P23.2
(a) 2.62 1024
P23.4
1.57 µN to the left
P23.6
2.51 10−9
P23.8
x = 0.634d. The equilibrium is stable if the third bead has positive charge.
P23.10
(a) period =
P23.12
1.82 m to the left of the negative charge
P23.14
514 kN
P23.16
(a) 12.9 ĵ kN C
P23.18
See the solution.
P23.20
ke λ 0
− î
x0
P23.22
See the solution.
P23.24
(a) 383 MN/C away
P23.26
See the solution.
π
2
(b) 2.38 electrons for every 109 present
md 3
where m is the mass of the object with charge −Q
ke qQ
(b) 4 a
ke qQ
md 3
(b) −38.6 ĵ mN
( )
(b) 324 MN/C away
(c) 80.7 MN/C away
(d) 6.68 MN/C away
25
26
Chapter 23
keQî ⎡ 2
( d + R2
h ⎢⎣
P23.28
(a)
P23.30
(a) 200 pC
P23.32
See the solution.
P23.34
(a) −
1
3
)
−1 2
(
− ( d + h ) + R2
(b) 141 pC
2
)
−1 2
⎤
⎥⎦
(b)
2 keQî ⎡
h + ( d 2 + R2
R 2 h ⎢⎣
)
12
(
− ( d + h ) + R2
2
)
12
⎤
⎥⎦
(c) 58.9 pC
(b) q1 is negative and q2 is positive
P23.36
(a) −57.6 î Tm s 2
P23.38
The particle strikes the negative plate after moving in a parabola 0.181 mm high and 0.961 mm wide.
P23.40
(a) 36.9°, 53.1° (b) 167 ns, 221 ns
P23.42
It is possible in just one way: at x = −16.0 cm place a charge of +51.3 µC.
P23.44
(a) 24.2 N/C at 0°
P23.46
(a)
mg
A cot θ + B
(b) 2.84 î Mm s
(b) 9.42 N/C at 117°
(b)
mgA
A cos θ + B sin θ
P23.48
0.205 µC
P23.50
ke qQ
toward the 29th vertex
a2
P23.52
25.9 cm
P23.54
(a) E =
P23.56
(c) 49.3 ns
935 N ⋅ m 2 x ˆi
(b) 4.00 î kN/C (c) At x = 0.016 8 m and at x = 0.916 m
3/ 2
C ( 0.0625 m 2 + x 2 )
(d) Nowhere is the field so large. (e) The field of a single charge in Question 7 takes on all
values from zero to infinity, each at just one point along the positive x-axis. The vector sum of
the field of two charges, in this problem, is zero at the origin, rises to a maximum at 17.7 cm, and
then decreases asymptotically to zero. In the question and the problem, the fields at x = 36 cm
happen to take similar values. For large x the field of the two charges in this problem shows
just the same inverse proportionality to x2 as the field in the question, being larger by the
factor 2(52 nC)/(57.7 nC) = 1.8 times.
⎛ mg ⎞
R⎜
⎟
⎝ ke 3 ⎠
12
P23.58
(a) At A downward. At B zero. At C upward. At E downward. At F downward.
solution. (c) 0.373 cm
P23.60
(a) See the solution.
P23.62
( −1.36ˆi + 1.96ˆj) kN C
P23.64
−
π 2 ke q
î
6a2
(b) k̂.
(b) See the
24
Gauss’s Law
CHAPTER OUTLINE
24.1
24.2
24.3
24.4
Electric Flux
Gauss’s Law
Application of Gauss’s Law to
Various Charge Distributions
Conductors in Electrostatic
Equilibrium
ANSWERS TO QUESTIONS
Q24.1
The luminous flux on a given area is less when the sun is
low in the sky, because the angle between the rays of the
sun and the local area vector, d A , is greater than zero.
The cosine of this angle is reduced. The decreased flux
results, on the average, in colder weather.
Q24.2
The surface must enclose a positive total charge.
Q24.3
The net flux through any gaussian surface is zero. We
can argue it two ways. Any surface contains zero charge,
so Gauss’s law says the total flux is zero. The field is
uniform, so the field lines entering one side of the closed
surface come out the other side and the net flux is zero.
*Q24.4
(i) Equal amounts of flux pass through each of the six faces of the cube. Answer (e).
(ii) Move the charge to very close below the center of one face, through which the flux is then
q/2∈0. Answer (c).
(iii) Move the charge onto one of the cube faces. Then the field has no component perpendicular
to this face and the flux is zero. Answer (a).
*Q24.5
(i) Answer (a).
(ii) the flux is zero through the two faces pierced by the filament. Answer (b).
*Q24.6
(i) Answer (a).
(ii) The flux is nonzero through the top and bottom faces, and zero through the other four faces.
Answer (c).
*Q24.7
(i) Both spheres create equal fields at exterior points, like particles at the centers of the spheres.
Answer (c).
(ii) The field within the conductor is zero. The field within the insulator is 4/5 of its surface
value. Answer (f).
Q24.8
Gauss’s law cannot tell the different values of the electric field at different points on the surface.
When E is an unknown number, then we can say
∫ E cosθ dA = E ∫ cosθ dA. When E ( x, y, z )
is an unknown function, then there is no such simplification.
Q24.9
The electric flux through a sphere around a point charge is independent of the size of the sphere.
A sphere of larger radius has a larger area, but a smaller field at its surface, so that the product
of field strength and area is independent of radius. If the surface is not spherical, some parts are
closer to the charge than others. In this case as well, smaller projected areas go with stronger
fields, so that the net flux is unaffected.
27
28
Chapter 24
Q24.10 Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels
every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer
surface of the conductor.
*Q24.11 (a)
(b)
Let q represent the charge of the insulating sphere. The field at A is (4/5)3q/[4p (4 cm)2∈0].
The field at B is q/[4p (8 cm)2∈0]. The field at C is zero. The field at D is q/[4p (16 cm)2∈0].
The ranking is A > B > D > C.
The flux through the 4-cm sphere is (4/5)3q/∈0. The flux through the 8-cm sphere and
through the 16-cm sphere is q/∈0. The flux through the 12-cm sphere is 0. The ranking is
B = D > A > C.
*Q24.12 The outer wall of the conducting shell will become polarized to cancel out the external field. The
interior field is the same as before. Answer (c).
Q24.13 If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shell
carries no charge. The person is not harmed by touching this wall. If the person carries a (small)
charge q, the electric field inside the sphere is no longer zero. Charge –q is induced on the inner
wall of the sphere. The person will get a (small) shock when touching the sphere, as all the charge
on his body jumps to the metal.
*Q24.14 (i) The shell becomes polarized. Answer (e).
(ii) The net charge on the shell’s inner and outer surfaces is zero. Answer (a).
(iii) Answer (c).
(iv) Answer (c).
(v) Answer (a).
Q24.15 There is zero force. The huge charged sheet creates a uniform field. The field can polarize the
neutral sheet, creating in effect a film of opposite charge on the near face and a film with an equal
amount of like charge on the far face of the neutral sheet. Since the field is uniform, the films of
charge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces
add to zero.
SOLUTIONS TO PROBLEMS
Section 24.1
P24.1
Electric Flux
Φ E = EA cos θ
A = π r 2 = π ( 0.200 ) = 0.126 m 2
5.20 × 10 5 = E ( 0.126 ) cos 0°
E = 4.14 × 10 6 N C = 4.14 MN C
2
P24.2
Φ E = EA cos θ = ( 2.00 × 10 4 N C ) (18.0 m 2 ) cos 10.0° = 355 kN ⋅ m 2 C
P24.3
(a)
Φ E = E ⋅ A = aˆi + bˆj ⋅ Aˆi = aA
(b)
ΦE
(c)
ΦE
( )
= ( aˆi + bˆj) ⋅ Aˆj = bA
= ( aˆi + bˆj) ⋅ Akˆ = 0
Gauss’s Law
P24.4
(a)
29
A ′ = (10.0 cm ) ( 30.0 cm )
A ′ = 300 cm 2 = 0.030 0 m 2
Φ E , A′ = EA ′ cos θ
)
Φ E , A′ = ( 7.80 × 10 4 ( 0.030 0 ) cos 180°
Φ E , A′ = −2.34 kN ⋅ m 2 C
(b)
)
Φ E , A = EA cos θ = ( 7.80 × 10 4 ( A ) cos 60.0°
FIG. P24.4
⎛ 10.0 cm ⎞
A = ( 30.0 cm ) ( w ) = ( 30.0 cm ) ⎜
= 600 cm 2 = 0.060 0 m 2
⎝ cos 60.0° ⎟⎠
)
Φ E , A = ( 7.80 × 10 4 ( 0.060 0 ) cos 60.0° = +2.34 kN ⋅ m 2 C
(c)
The bottom and the two triangular sides all lie parallel to E, so Φ E = 0 for each of these.
Thus,
Φ E,total = −2.34 kN ⋅ m 2 C + 2.34 kN ⋅ m 2 C + 0 + 0 + 0 = 0
P24.5
Φ E = EA cos θ through the base
Φ E = ( 52.0 ) ( 36.0 ) cos 180° = −1.87 kN ⋅ m 2 C
Note that the same number of electric field lines go through the base as go
through the pyramid’s surface (not counting the base).
FIG. P24.5
For the slanting surfaces, Φ E = +1.87 kN ⋅ m 2 C .
Section 24.2
P24.6
(a)
Gauss’s Law
E=
keQ
;
r2
But Q is negative since E points inward.
P24.7
(8.99 × 10 ) Q
9
8.90 × 10 2 =
( 0.750 )2
Q = −5.57 × 10 −8 C = −55.7 nC
(b)
The negative charge has a spherically symmetric
the spherical shell.
(a)
ΦE =
charge distribution, concentric with
qin ( +5.00 µC − 9.00 µC + 27.0 µC − 84.0 µC )
=
= −6.89 × 10 6 N ⋅ m 2 C2
−12
2
2
∈0
8.85 × 10 C N ⋅ m
Φ E = −6.89 MN ⋅ m 2 C
(b)
Since the net electric flux is negative, more lines enter than leave the surface.
30
P24.8
Chapter 24
(a)
One-half of the total flux created by the charge q goes through the plane. Thus,
Φ E ,plane =
(b)
1
1⎛ q ⎞
q
Φ E ,total = ⎜ ⎟ =
2
2 ⎝ ∈0 ⎠
2∈0
The square looks like an infinite plane to a charge very close to the surface. Hence,
Φ E , square ≈ Φ E ,plane =
(c)
P24.9
P24.10
P24.11
ΦE =
q
2∈0
The plane and the square look the same to thhe charge.
qin
∈0
Through S1
ΦE =
−2Q + Q
Q
= −
∈0
∈0
Through S2
ΦE =
+Q − Q
= 0
∈0
Through S3
ΦE =
−2Q + Q − Q
2Q
= −
∈0
∈0
Through S4
ΦE = 0
qin 12.0 × 10 −6
=
= 1.36 × 100 6 N ⋅ m 2 C = 1.36 MN ⋅ m 2 C
∈0 8.85 × 10 −12
1
= (1.36 × 10 6 N ⋅ m 2 C = 6.78 × 10 5 N ⋅ m 2 C = 678 kN ⋅ m 2 C
2
(a)
Φ E , shell =
(b)
Φ E,half shell
(c)
No , the same number of field lines will pass through each surface, no matter how the
radius changes.
(a)
With d very small, all points on the hemisphere are
nearly at a distance R from the charge, so the field
kQ
everywhere on the curved surface is e 2 radially
R
outward (normal to the surface). Therefore, the flux
is this field strength times the area of half a sphere:
Φ curved = ∫ E ⋅ dA = Elocal Ahemisphere
)
δ→0
Q 1
1
+Q
Q ( 2π ) =
Φ curved = ⎛ ke 2 ⎞ ⎛ 4π R 2 ⎞ =
⎝ R ⎠⎝2
⎠ 4π ∈0
2∈0
(b)
FIG. P24.11
The closed surface encloses zero charge so Gauss’s law gives
Φ curved + Φ flat = 0
P24.12
Q
or
Φ flat = − Φ curved =
−Q
2∈0
Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.
ρ A (100 m )
q
∫ E ⋅ dA = ∈ : ( +120 N C) A + ( −100 N C) A = ∈
0
0
ρ=
( 20 N C ) (8.85 × 10 −12 C2
100 m
N ⋅ m2 )
= 1.77 × 10 −12 C m 3
The charge is positive , to produce the net outward flux of electric field.
Gauss’s Law
P24.13
The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q , of which
∈0
one-sixth goes through each face:
( Φ E )one face =
Q−6 q
6∈0
( Φ E )one face =
Q − 6 q ( 5.00 − 6.00 ) × 10 −6 C ⋅ N ⋅ m 2
= −18.8 kN ⋅ m 2 C
=
6∈0
6 × 8.85 × 10 −12 C2
.
P24.14
The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q , of which
∈0
one-sixth goes through each face:
( Φ E )one face =
P24.15
Q−6 q
6∈0
If R ≤ d , the sphere encloses no charge and Φ E =
qin
= 0 .
∈0
If R > d , the length of line falling within the sphere is 2 R 2 − d 2
so
P24.16
ΦE =
2λ R 2 − d 2
∈0
⎛ (8.99 × 10 9 N ⋅ m 2 C2 ) (10.0 × 10 −6 C ) ⎞
2
kQ
Φ E ,hole = E ⋅ A hole = ⎛ e 2 ⎞ (π r 2 ) = ⎜
π (1.00 × 10 −3 m )
2
⎟
⎝ R ⎠
)
( 0.100 m
⎝
⎠
Φ E ,hole = 28.2 N ⋅ m 2 C
P24.17
ΦE =
qin
170 × 10 −6 C
=
= 1.92 × 10 7 N ⋅ m 2 /C
∈0 8.85 × 10 −12 C2 N ⋅ m 2
(a)
( Φ E )one face =
(b)
Φ E = 19.2 MN ⋅ m 2 C
(c)
The answer to (a) would change because the flux through each face of the cube would
nott be equal with an asymmetric charge distriibution. The sides of the cube nearer the
chharge would have more flux and the ones furrther away would have less. The answer
to (bb) would remain the same, since the overalll flux would remain the same.
1
1.92 × 10 7 N ⋅ m 2 C
ΦE =
6
6
( Φ E )one face =
3.20 MN ⋅ m 2 C
31
32
Chapter 24
Section 24.3
P24.18
(a)
(b)
(c)
(d)
Application of Gauss’s Law to Various Charge Distributions
keQr
= 0
a3
9
−6
k Qr (8.99 × 10 ) ( 26.0 × 10 ) ( 0.100 )
E= e3 =
= 365 kN C
a
( 0.400 )3
9
−6
k Q (8.99 × 10 ) ( 26.0 × 10 )
E = e2 =
= 1.46 MN
NC
r
( 0.400 )2
E=
9
−6
keQ (8.99 × 10 ) ( 26.0 × 10 )
=
= 649 kN C
2
2
r
( 0.600 )
E=
The direction for each electric field is radially outward .
P24.19
The charge distributed through the nucleus creates a field at the surface equal to that of a point
kq
charge at its center: E = e2 .
r
E=
(8.99 × 10
9
Nm 2 C2 ) (82 × 1.60 × 10 −19 C )
⎡⎣( 208 )1 3 1.20 × 10 −15 m ⎤⎦
E = 2.33 × 10 21 N C
P24.20
away from the nucleus
Note that the electric field in each case is directed radially inward, toward the filament.
(a)
(b)
(c)
P24.21
E=
P24.22
(a)
(b)
P24.23
2
)
)
)
)
)
)
9
2
2
−6
2 ke λ 2 ( 8.99 × 10 N ⋅ m C ( 90.0 × 10 C m
=
= 16.2 MN C
0.100 m
r
9
2
2
−6
2 k λ 2 ( 8.99 × 10 N ⋅ m C ( 90.0 × 10 C m
= 8.09 MN C
E= e =
0.200 m
r
9
2
2
−6
2 k λ 2 ( 8.99 × 10 N ⋅ m C ( 90.0 × 10 C m
E= e =
= 1.62 MN C
1.00 m
r
E=
σ
9.00 × 10 −6 C m 2
=
= 508 kN C, upward
2 ∈0 2 ( 8.85 × 10 −12 C2 N ⋅ m 2 )
2 (8.99 × 10 9 ) (Q 2.40 )
2 ke λ
4
3
.
60
×
10
=
E=
0.190
r
−7
Q = +9.13 × 10 C = +913 nC
E= 0
⎛ Q A⎞
⎛ σ ⎞
= q⎜
mg = qE = q ⎜
⎟
⎝ 2 ∈0 ⎟⎠
⎝ 2 ∈0 ⎠
−12
Q 2∈0 mg 2 (8.85 × 10 ) ( 0.01) ( 9.8 )
=
=
A
q
−0.7 × 10 −6
= −2.48 µ C m 2
*P24.24 (a)
(b)
A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributed
throughout its volume, with density 5.00 µC/m3. Find the magnitude of the electric field
it creates at a point P, 3.00 cm from its axis. As a gaussian surface choose a concentric
cylinder with its curved surface passing through the point P and with length 8.00 cm.
We solve for
(0.02 m)2 0.08 m (5 × 10 −6 C/m 3 )
E=
= 3.77 kN/C
(8.85 × 10 −12 C2 /N ⋅ m 2 )2(0.03 m) 0.08 m
Gauss’s Law
P24.25
The volume of the spherical shell is
4
π ⎡( 0.25 m )3 − ( 0.20 m )3 ⎤⎦ = 3.19 × 10 −2 m 3
3 ⎣
Its charge is
ρV = ( −1.33 × 10 −6 C m 3 ) ( 3.19 × 10 −2 m 3 ) = −4.25 × 10 −8 C
The net charge inside a sphere containing the proton’s path as its equator is
−60 × 10 −9 C − 4.25 × 10 −8 C = −1.02 × 10 −7 C
The electric field is radially inward with magnitude
)
8.99 × 10 9 Nm 2 (1.02 × 10 −7 C
ke q
q
= 1.47 × 10 4 N C
=
=
2
r2
∈0 4π r 2
C2 ( 0.25 m )
For the proton
∑ F = ma
eEr ⎞
v=⎛
⎝ m ⎠
eE =
12
mv 2
r
⎛ 1.60 × 10 −19 C (1.47 × 10 4 N C ) 0.25 m ⎞
=⎜
⎟
1.67 × 10 −27 kg
⎝
⎠
12
= 5.94 × 10 5 m s
2
⎛ 100 cm ⎞
= 8.60 × 10 −2 C m 2
*P24.26 s = ( 8.60 × 10 −6 C cm 2 ⎜
⎝ m ⎟⎠
)
E=
σ
8.60 × 10 −2
=
= 4.86 × 10 9 N C awaay from the wall
2 ∈0 2 ( 8.85 × 10 −12 )
So long as the distance from the wall is small compared to the width and height of the wall, the
distance does not affect the field.
P24.27
If r is positive, the field must be radially outward. Choose as the
gaussian surface a cylinder of length L and radius r, contained
inside the charged rod. Its volume is π r 2 L and it encloses
charge ρπ r 2 L. Because the charge distribution
is long, no electric
E
⋅ dA = EdA cos 90.0° = 0 .
flux passes through the circular
end
caps;
The curved surface has E ⋅ dA = EdA cos 0°, and E must be the same
strength everywhere over the curved surface.
Gauss’s law,
q
∫ E ⋅ dA = ∈ ,
becomes
E
0
∫
dA =
Curved
Surface
FIG. P24.27
ρπ r 2 L
.
∈0
Now the lateral surface area of the cylinder is 2π rL :
E ( 2π r ) L =
P24.28
ρπ r 2 L
∈0
Thus,
E=
ρr
radially away from the cylinder axiis .
2 ∈0
The distance between centers is 2 × 5.90 × 10−15 m. Each produces a field as if it were a point
charge at its center, and each feels a force as if all its charge were a point at its center.
( 46 ) (1.60 × 10 −19 C )
ke q1q2
3
9
2
2
=
8
.
99
×
10
N
⋅
m
C
(
)
2 = 3.50 × 10 N = 3.50 kN
−15
r2
( 2 × 5.90 × 10 m )
2
F=
2
33
34
P24.29
P24.30
Chapter 24
(a)
E= 0
(b)
E=
9
−6
keQ (8.99 × 10 ) ( 32.0 × 10 )
=
= 7.19 MN C
r2
( 0.200 )2
E = 7.19 MN C radially outward
Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with
a 0.05 m separation on the ends of strings making angles of 10° with the vertical.
(a)
mg
cos 10°
∑F
= T cos 10° − mg = 0 ⇒ T =
∑F
= T sin 10° − Fe = 0 ⇒ Fe = T sin 10°, so
y
x
FIG. P24.30
mg ⎞
Fe = ⎛
sin 10° = mg tan 10° = ( 0.001 kg ) ( 9.8 m s 2 ) tan 10°
⎝ cos 10° ⎠
Fe ≈ 2 × 10 −3 N ~10 −3 N or 1 mN
(b)
Fe =
ke q 2
r2
2 × 10 −3 N ≈
(8.99 × 10
9
N ⋅ m 2 C2 ) q 2
( 0.25 m )2
q ≈ 1.2 × 10 −7 C ~10 −7 C or 100 nC
P24.31
E=
(d)
ΦE =
(a)
)
)
9
2
2
−7
ke q ( 8.99 × 10 N ⋅ m C (1.2 × 10 C
≈
≈ 1.7 × 10 4 N C ~ 10 kN C
r2
( 0.25 m )2
(c)
q
1.2 × 10 −7 C
≈
= 1.4 × 10 4 N ⋅ m 2 C ~10 kN ⋅ m 2 C
∈0 8.85 × 10 −12 C2 N ⋅ m 2
9
2
2
−6
2 ke λ 2 (8.99 × 10 N ⋅ m C ) ⎡⎣( 2.00 × 10 C ) 7.00 m ⎤⎦
E=
=
0.100 m
r
E = 51.4 kN C , radially outward
(b)
Φ E = EA cos θ = E ( 2π r ) cos 0°
Φ E = ( 5.14 × 10 4 N C ) 2π ( 0.100 m ) ( 0.020 0 m ) (1.00 ) = 646 N ⋅ m 2 C
Section 24.4
P24.32
Conductors in Electrostatic Equilibrium
σ conductor
suggested in the chapter for the field outside the
∈0
σ
aluminum looks different from the equation E = insulator for the field around glass. But its charge
2∈0
Q
.
will spread out to cover both sides of the aluminum plate, so the density is σ conductor =
2A
Q
Q
The glass carries charge only on area A, with σ insulator = . The two fields are
, the same
2 A∈0
A
The fields are equal. The equation E =
in magnitude, and both are perpendicular to the plates, vertically upward if Q is positive.
Gauss’s Law
P24.33
qin
∫ EdA = E ( 2π rl ) = ∈
0
E=
λ
qin l
=
2π ∈0 r 2π ∈0 r
(a)
r = 3.00 cm
E= 0
(b)
r = 10.0 cm
E=
30.0 × 10 −9
= 5 400 N C, outward
2π (8.85 × 10 −12 ) ( 0.100 )
(c)
r = 100 cm
E=
30.0 × 10 −9
= 540 N C, outtward
2π (8.85 × 10 −12 ) (1.00 )
*P24.34 (a)
35
All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside
is zero .
(b)
The charged sphere creates field at exterior points as if it were a point charge at the center:
kq
(8.99 × 109 N·m 2 ) ( 40 × 10 −9 C) outward = 1.24 × 10 4 N C outward
E = e2 away =
2
r
C2 ( 0.17 m )
(c)
P24.35
(8.99 × 10 9 N·m 2 ) ( 40 × 10 −9 C )
outward = 639 N C outward
E=
2
C2 ( 0.75 m )
(d)
All three answers would be the same. The solid copper sphere carries charge only on its
outer surface.
(a)
E=
σ
∈0
σ = ( 8.00 × 10 4 ) ( 8.85 × 10 −12 ) = 7.08 × 10 −7 C m 2
σ = 708 nC m 2 , positive on one face and negative on the other.
(b)
σ=
Q
A
Q = σ A = ( 7.08 × 10 −7 ) ( 0.500 ) C
2
Q = 1.77 × 10 −7 C = 177 nC , positive on one face and negative on the other.
*P24.36 Let the flat box have face area A perpendicular to its thickness dx. The flux at x = 0.3 m is into
the box −EA = −(6 000 N/C ⋅ m2)(0.3 m)2 A = −(540 N/C) A
The flux out of the box at x = 0.3 m + dx
+EA = −(6 000 N/C ⋅ m2)(0.3 m + dx)2 A = +(540 N/C) A + (3 600 N/C ⋅ m) dx A
(The term in (dx)2 is negligible.)
The charge in the box is rA dx where r is the unknown. Gauss’s law is
−(540 N/C) A + (540 N/C) A + (3 600 N/C ⋅ m) dx A = rA dx/∈0
Then r = (3600 N/C ⋅ m)∈0 = (3600 N/C ⋅ m)(8.85 × 10−12 C2/N ⋅ m2) = 31.9 nC/m3
P24.37
Q
The charge divides equally between the identical spheres, with charge
on each. Then they
2
repel like point charges at their centers:
8.99 × 10 9 N ⋅ m 2 ( 60.0 × 10 −6 C )
ke ( Q 2 ) ( Q 2 )
keQ 2
=
=
= 2.00 N
2
4 C2 ( 2.01 m )
( L + R + R )2 4 ( L + 2 R )2
2
F=
36
Chapter 24
*P24.38 The surface area is A = 4pa2. The field is then
kQ
Q
Q
σ
E = e2 =
=
=
2
a
A ∈0 ∈0
4π ∈0 a
It is not equal to s /2∈0. At a point just outside, the uniformly charged surface looks just like a
uniform flat sheet of charge. The distance to the field point is negligible compared to the radius
of curvature of the surface.
P24.39
(a)
Inside surface: consider a cylindrical surface within the metal. Since E inside the
conducting shell is zero, the total charge inside the gaussian surface must be zero, so the
inside charge/length = − λ .
qin
0 = λ + qin
so
= −λ
2λ = qin + qout
Outside surface: The total charge on the metal cylinder is
qout = 2λ + λ (b)
P24.40
E=
3λ
so the outside charge/length is
2 ke ( 3λ ) 6 ke λ
3λ
=
=
radially outward
r
r
2π ∈0 r
An approximate sketch is given at the right. Note that the electric field
lines should be perpendicular to the conductor both inside and outside.
FIG. P24.40
P24.41
(a)
The charge density on each of the surfaces (upper and lower) of the plate is:
−8
1 q
1 ( 4.00 × 10 C )
σ= ⎛ ⎞=
= 8.00 × 10 −8 C m 2 = 80.0 nC m 2
2
2 ⎝ A ⎠ 2 ( 0.500 m )
(b)
⎛σ⎞
⎛ 8.00 × 10 −8 C m 2 ⎞ ˆ
E = ⎜ ⎟ kˆ = ⎜
k=
⎝ ∈0 ⎠
⎝ 8.85 × 10 −12 C2 N ⋅ m 2 ⎟⎠
(c)
E=
( −9.04
( 9.04
kN C ) kˆ
kN C ) kˆ
Additional Problems
P24.42
E = ayˆi + bzˆj + cxkˆ
In general,
In the xy plane, z = 0
and
w
)
2 w
x
Φ E = ch ∫ xdx = ch
2
x=0
x=0
y=0
E = ayˆi + cxkˆ
Φ E = ∫ E ⋅ dA = ∫ ayˆi + cxkˆ ⋅ kˆ dA
(
z
chw
=
2
x=0
y
x=w
2
y=h
dA = hdx
x
FIG. P24.42
Gauss’s Law
P24.43
(a)
37
Uniform E, pointing radially outward, so Φ E = EA . The arc length
is ds = Rdθ , and the circumference is 2π r = 2π R sin θ .
θ
θ
A = ∫ 2π rds = ∫ ( 2π R sin θ ) Rdθ = 2π R 2 ∫ sin θ dθ
0
FIG. P24.43
0
= 2π R ( − cos θ ) 0 = 2π R (1 − cos θ )
θ
2
ΦE =
2
1 Q
Q
⋅ 2π R 2 (1 − cos θ ) =
(1 − cosθ ) [independent of R!]
2
4π ∈0 R
2 ∈0
(b)
For θ = 90.0° (hemisphere): Φ E =
Q
Q
.
(1 − cos 90°) =
2 ∈0
2 ∈0
(c)
For θ = 180° (entire sphere): Φ E =
Q
Q
(1 − cos180°) =
2 ∈0
∈0
*P24.44 (a)
[Gauss’s Law].
qin = +3 µ C − 1 µ C = +2.00 µ C
(b)
The charge distribution is spherically symmetric and qin > 0. Thus, the field is directed
radially outward or to the right at point D.
(c)
E=
(d)
Since all points within this region are located inside conducting material, E = 0 .
(e)
Φ E = ∫ E ⋅ dA = 0
(f )
qin = +3.00 µ C
(g)
E=
(h)
⎛ +3 µ C ⎞ 4
qin = ρV = ⎜ 4 3 ⎟ ⎛ π 4 3 ⎞ = +1.54 µ C
⎠
⎝ 3π5 ⎠⎝ 3
(i)
E=
(j)
As in part (d), E = 0 for 10 cm < r < 15 cm .
E
Thus, for a spherical gaussian surface with
10 cm < r < 15 cm, qin = +3 µ C + qinner = 0
where qinner is the charge on the inner surface
ke qin 8.99 × 10 9 2.00 × 10 −6 N/C
=
= 702 kN/C
r2
(0.16)2
⇒
qin = ∈0 Φ E = 0
ke qin 8.99 × 10 9 3.00 × 10 −6
=
= 4.21 MN/C to the right (radially outward).
r2
(0.08 )2
ke qin 8.99 × 10 9 1.54 × 10 −6
=
= 8.63 MN/C to the right (radially outward)
r2
(0.04 )2
of the conducting shell. This yields
qinner = −3.00 µ C .
(k)
Since the total charge on the conducting
shell is qnet = qouter + qinner = −1 µ C, we have
qouter = −1 µ C − qinner = −1 µ C − ( −3 µ C ) = +2.00 µ C
(l)
This is shown in the figure to the right.
a
b
c
FIG. P24.44(l)
r
38
Chapter 24
*P24.45 (a)
The field is zero within the metal of the shell. The exterior electric field lines end at equally
spaced points on the outer surface. The charge on the outer surface is distributed uniformly.
Its amount is given by
EA = Q/∈0
Q = −(890 N/C) 4p (0.75 m)2 8.85 × 10−12 C2/N ⋅ m2 = −55.7 nC
(b) and (c)
For the net charge of the shell to be zero, the shell must carry +55.7 nC on its inner
surface, induced there by −55.7 nC in the cavity within the shell. The charge in the
cavity could have any distribution and give any corresponding distribution to the
charge on the inner surface of the shell. For example, a large positive charge might
be within the cavity close to its topmost point, and a slightly larger negative charge
near its easternmost point. The inner surface of the shell would then have plenty
of negative charge near the top and even more positive charge centered on the
eastern side.
P24.46
The sphere with large charge creates a strong field to polarize the other sphere. That means it
pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side.
This patch of opposite charge is smaller in amount but located in a stronger external field, so it
can feel a force of attraction that is larger than the repelling force felt by the larger charge in the
weaker field on the other side.
P24.47
(a)
∫ E ⋅ dA = E ( 4π r ) = ∈
2
qin
0
(b)
For r < a,
4
qin = ρ ⎛ π r 3 ⎞
⎝3
⎠
so
E=
For a < r < b and c < r,
qin = Q
So
E=
For b ≤ r ≤ c,
E = 0,
ρr
3 ∈0
FIG. P24.47
Q
4π r 2 ∈0
since E = 0 inside a conductor.
Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the
conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.
Therefore,
q1 + Q = 0
and
σ1 =
q1
−Q
=
2
4π b
4π b 2
Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow
sphere is uncharged, we require
q1 + q2 = 0
and
σ2 =
q1
Q
=
4π c 2
4π c 2
Gauss’s Law
P24.48
First, consider the field at distance r < R from the center of a uniform sphere of positive charge
(Q = + e) with radius R.
( 4π r ) E = q∈
2
=
in
0
ρV ⎛ + e ⎞ 43 π r 3
=
∈0 ⎜⎝ 43 π R 3 ⎟⎠ ∈0
⎛
⎞
e
r directed outward
E=⎜
3 ⎟
⎝ 4π ∈0 R ⎠
so
The force exerted on a point charge q = − e located at distance r from the center is then
(a)
⎛
e
F = qE = − e ⎜
⎝ 4π ∈0 R3
⎛
⎞
⎞
e2
r
=
−
⎜⎝ 4π ∈ R3 ⎟⎠ r = − Kr
⎟⎠
0
e2
k e2
= e3
3
R
4π ∈0 R
(b)
K=
(c)
⎛ k e2 ⎞
Fr = me ar = − ⎜ e 3 ⎟ r,
⎝ R ⎠
⎛ k e2 ⎞
ar = − ⎜ e 3 ⎟ r = −ω 2 r
⎝ me R ⎠
so
f=
Thus, the motion is simple harmonic with frequency
f = 2.47 × 1015 Hz =
(d)
( 8.99 × 10 N ⋅ m C ) (1.60 × 10
( 9.11 × 10 kg ) R
9
1
2π
2
2
−31
which yields R 3 = 1.05 × 10 −30 m 3 ,
P24.49
39
or
ω
1
=
2π
2π
−19
d vy
dt
Now
= Fy
m
2
R = 1.02 × 10 −10 m = 102 pm
d vy dx
= qEy
dx dt
y
v
q
0
vy
vx x
d
Q
dx
= vx has the nearly constant value v. So
dt
vy
)
3
The vertical velocity component of the moving charge
increases according to
m
C
ke e 2
me R 3
FIG. P24.49
⬁
q
vy = ∫ d vy =
Ey dx
m v −∫⬁
0
q
d vy =
Ey dx
mv
The radially outward component of the electric field varies along the x axis, but is described by
⬁
∫
⬁
Ey dA =
−⬁
Q
∫ E ( 2π d ) dx = ∈
y
−⬁
⬁
So
0
Q
∫ E dx = 2π d ∈
y
−⬁
tan θ =
and vy =
0
vy
v
=
qQ
2π ∈0 dm v 2
qQ
. The angle of deflection is described by
m v 2π d ∈0
θ = tan −1
qQ
2π ∈0 dmv 2
40
P24.50
Chapter 24
Consider the field due to a single sheet and let E+ and E− represent
the fields due to the positive and negative sheets. The field at any
distance from each sheet has a magnitude given by the textbook equation
E+ = E− =
σ
2 ∈0
(a)
To the left of the positive sheet, E+ is directed toward the
left and E− toward the right and the net field over this
region is E = 0 .
(b)
In the region between the sheets, E+ and E− are both directed
toward the right and the net field is
σ
E=
to the right
∈0
(c)
P24.51
To the right of the negative sheet, E+ and E− are again
oppositely directed and E = 0 .
FIG. P24.50
The magnitude of the field due to each sheet given by
Equation 24.8 is
E=
(a)
σ
directed perpendicular to the sheet
2 ∈0
In the region to the left of the pair of sheets, both fields are
directed toward the left and the net field is
σ
E=
to the left
∈0
FIG. P24.51
(b)
In the region between the sheets, the fields due to the individual sheets are oppositely
directed and the net field is
E= 0
(c)
In the region to the right of the pair of sheets, both fields are directed toward the right and
the net field is
σ
E=
to the right
∈0
P24.52
The resultant field
within the cavity is the superposition of
two fields, one E+ due to a uniform
sphere of positive charge
of radius 2a, and the other E− due to a sphere of negative
charge of radius a centered within the cavity.
4 ⎛ π r 3ρ ⎞
= 4 π r 2 E+
3 ⎜⎝ ∈0 ⎟⎠
so
ρr
ρr
E+ =
rˆ =
3∈0
3∈0
4 ⎛ π r 3ρ ⎞
− ⎜ 1 ⎟ = 4π r12 E−
3 ⎝ ∈0 ⎠
so
ρr
−ρ E− = 1 ( − rˆ1 ) =
r1
3∈0
3∈0
FIG. P24.52
continued on next page
Gauss’s Law
−ρ (r − a)
=
E−
3∈0
Since r = a + r1 ,
ρr
ρr
ρa
ρa
ρa
E = E+ + E− =
−
+
=
= 0 ˆi +
ĵ
3∈0 3∈0 3∈0 3∈0
3∈0
P24.53
Thus,
Ex = 0
and
Ey =
at all points within the cavity
Consider the charge distribution to be an unbroken charged spherical shell with uniform charge
density s and a circular disk with charge per area −σ . The total field is that due to the whole
Q
4π R 2σ
σ
σ
σ
=
outward plus the field of the disk −
radially
sphere,
=
=
2
2
4π ∈0 R
4π ∈0 R
∈0
2 ∈0 2 ∈0
inward. The total field is
P24.54
ρa
3∈0
σ
σ
σ
−
=
outward .
∈0 2 ∈0
2 ∈0
The
electric field throughout the region is directed along x; therefore,
E will be perpendicular to dA over the four faces of the surface which are
perpendicular to the yz plane, and E will be parallel to dA over the two faces
which are parallel to the yz plane. Therefore,
(
Φ E = − Ex
x=a
) A + (E
) A = − (3 + 2a ) ab + (3 + 2 ( a + c ) ) ab
2
2
x x= a+ c
= 2abc ( 2a + c )
FIG. P24.54
Substituting the given values for a, b, and c, we find Φ E = 0.269 N ⋅ m 2 C .
Q = ∈ 0 Φ E = 2.38 × 10 −12 C = 2.38 pC
P24.55
∫ E ⋅ dA = E ( 4π r ) = ∈
2
qin
0
R
(a)
For r > R,
qin = ∫ Ar 2 ( 4π r 2 ) dr = 4π
0
E=
and
AR 5
5 ∈0 r 2
r
(b)
For r < R,
qin = ∫ Ar 2 ( 4π r 2 ) dr =
0
and
E=
AR5
5
Ar 3
5 ∈0
4π Ar 5
5
41
42
P24.56
Chapter 24
Q
.
The total flux through a surface enclosing the charge Q is
∈
0
The flux through the disk is
Φ disk = ∫ E ⋅ dA
where the integration covers the area of the disk. We must evaluate this
1
Q
integral and set it equal to 4
to find how b and R are related. In the figure,
∈
0
take d A to be the area of an annular ring of radius s and width ds. The flux
through d A is E ⋅ dA = EdA cos θ = E ( 2π sds ) cos θ .
FIG. P24.56
The magnitude of the electric field has the same value at all points within the annular ring,
E=
1 Q
1
Q
=
4π ∈0 r 2 4π ∈0 s 2 + b 2
and
cos θ =
b
b
=
r ( s2 + b2
)
12
.
Integrate from s = 0 to s = R to get the flux through the entire disk.
Φ E , disk =
Qb
2 ∈0
R
∫ (s
sds
2
0
+ b2 )
32
=
R
⎤
12
Qb ⎡
Q ⎡
b
⎢1 − 2
⎥
− ( s2 + b2 ) ⎤ =
1
2
⎦
2 ∈0 ⎣
2 ∈0 ⎢ ( R + b 2 ) ⎥
0
⎣
⎦
The flux through the disk equals
This is satisfied if R = 3b .
P24.57
qin
∫ E ⋅ dA = ∈
=
0
1
∈0
r
a
Q
b
provided that
2
4 ∈0
( R + b2
)
12
=
1
.
2
∫ r 4π r dr
2
0
4π a
4π a r 2
rdr
=
∈0 ∫0
∈0 2
r
E 4π r 2 =
E=
a
2 ∈0
= constant maagnitude
(The direction is radially outward from center for positive a; radially inward for negative a.)
P24.58
1
In this case the charge density is not uniform, and Gauss’s law is written as ∫ E ⋅ dA = ∈0 ∫ ρdV .
We use a gaussian surface which is a cylinder of radius r, length , and is coaxial with the charge
distribution.
(a)
When r < R, this becomes E ( 2π r ) =
ρ0 ⎛
r
a − ⎞ dV . The element of volume is a
∫
⎝
∈0 0
b⎠
r
cylindrical shell of radius r, length , and thickness dr so that dV = 2π r dr.
⎛ 2π r 2 ρ0 ⎞ ⎛ a r ⎞
ρ0 r
E ( 2π r ) = ⎜
⎜⎝ − ⎟⎠ so inside the cylinder, E =
⎟
2 ∈0
⎝ ∈0 ⎠ 2 3b
(b)
2r ⎞
⎛
⎜⎝ a − ⎟⎠
3b
When r > R , Gauss’s law becomes
E ( 2π r ) =
ρ0 R 2 ⎛
2R ⎞
ρ0 ⎛
r⎞
a
−
2
π
r
dr
E
or
outside
the
cylinder,
=
a−
(
)
⎟⎠
⎜⎝
∫
⎝
2 ∈0 r
3b ⎠
∈0 0
b
R
Gauss’s Law
P24.59
(a)
Consider a cylindrical shaped gaussian surface
perpendicular to the yz plane with one end in the yz plane
and the other end containing the point x:
Use Gauss’s law:
y
qin
∫ E ⋅ dA = ∈
gaussian
surface
0
By symmetry, the electricfield is zero in the yz plane
and is perpendicular to d A over the wall of the gaussian
cylinder. Therefore, the only contribution to the integral
is over the end cap containing the point x:
qin
∫ E ⋅ dA = ∈
or EA =
0
P24.60
a=
x
x
z
ρ ( Ax )
∈0
so that at distance x from the mid-line of the slab, E =
(b)
ρx
.
∈0
⎛ ρe ⎞
F ( − e) E
=
= −⎜
x
me
me
⎝ me ∈0 ⎟⎠
FIG. P24.59
The acceleration of the electron is of the form
a = −ω 2 x with ω =
Thus, the motion is simple harmonic with frequency
f=
ω
1
=
2π
2π
ρe
me ∈0
ρe
me ∈0
Consider the gaussian surface described in the solution to problem 59.
(a)
For x >
d
,
2
dq = ρ dV = ρ Adx = CAx 2 dx
1
∫ E ⋅ dA = ∈ ∫ dq
0
EA =
E=
(b)
CA
∈0
d 2
0
Cd 3
24 ∈0
For −
1 ⎛ CA ⎞ ⎛ d 3 ⎞
⎟⎜ ⎟
0 ⎠⎝ 8 ⎠
∫ x dx = 3 ⎜⎝ ∈
2
or
d
d
<x<
2
2
Cx 3
ˆi for x > 0;
E=
3∈0
43
Cd 3 ˆ
d
E=
i for x > ;
24 ∈0
2
1
∫ E ⋅ dA = ∈0
∫ dq =
Cx 3 ˆ
E=−
i for x < 0
3∈0
Cd 3 ˆ
d
E=−
i for x < −
24 ∈0
2
x
CA 2
CAx 3
x
dx
=
∈0 ∫0
3∈0
44
P24.61
Chapter 24
(a)
A point mass m creates a gravitational acceleration
Gm
g = − 2 rˆ at a distance r
r
Gm
2
∫ g ⋅ dA = − r 2 ( 4π r = −4π Gm
)
The flux of this field through a sphere is
Since the r has divided out, we can visualize the field as unbroken field lines. The same
flux would go through any other closed surface around the mass. If there are several or no
masses inside a closed surface, each creates field to make its own contribution to the net
flux according to
∫ g ⋅ dA = −4π Gmin
(b)
P24.62
Take a spherical gaussian surface of radius r. The field is inward so
2
2
∫ g ⋅ dA = g 4π r cos180° = − g 4π r
and
4
−4π Gmin = −4π G π r 3 ρ
3
Then,
4
4
− g 4π r 2 = −4π G π r 3 ρ and g = π r ρG
3
3
Or, since
ρ=
4
3
M Gr
ME
, g = E3
3
RE
π RE
The charge density is determined by Q =
(a)
M Gr
or g = E 3 inward
RE
4
π a3ρ
3
ρ=
3Q
4π a 3
The flux is that created by the enclosed charge within radius r:
ΦE =
(b)
ΦE =
(c)
ΦE
qin 4π r 3 ρ 4π r 3 3Q
Qr 3
=
=
=
3
∈0
3∈0
3∈0 4π a
∈0 a 3
Q
. Note that the answers to parts (a) and (b) agree at r = a .
∈0
Q
∈0
0
0
a
FIG. P24.62(c)
r
Gauss’s Law
P24.63
∫ E ⋅ dA = E ( 4π r ) = ∈
2
qin
0
( −3.60 × 10
(a)
3
)
N C 4π ( 0.100 m ) =
2
Q
8.85 × 10 −12 C2 N ⋅ m 2
(a < r < b)
Q = −4.00 × 10 −9 C = −4.00 nC
We take Q ′ to be the net charge on the hollow sphere. Outside c,
(b)
( +2.00 × 10
2
)
N C 4π ( 0.500 m ) =
2
Q + Q′
8.85 × 10 −12 C2 N ⋅ m 2
(r > c)
Q + Q ′ = +5.56 × 10 −9 C , so Q ′ = +9.56 × 10 −9 C = +9.56 nC
For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner
(c)
surface of the hollow sphere. Thus, Q1 = −Q = +4.00 nC .
Then, if Q2 is the total charge on the outer surface of the hollow sphere,
Q2 = Q′ − Q1 = 9.56 nC − 4.0 nC = +5.56 nC .
P24.64
The field direction is radially outward perpendicular to the axis. The field strength depends on
r but not on the other cylindrical coordinates q or z. Choose a gaussian cylinder of radius r and
length L. If r < a ,
ΦE =
E=
qin
∈0
λ
2π r ∈0
If a < r < b ,
and
or
E ( 2π rL ) =
E=
λL
∈0
λ
rˆ
2π r ∈0
E ( 2π rL ) =
(r < a)
λ L + ρπ ( r 2 − a 2 ) L
∈0
λ + ρπ ( r 2 − a 2 )
E=
rˆ
2π r ∈0
If r > b ,
E ( 2π rL ) =
λ L + ρπ ( b 2 − a 2 ) L
∈0
λ + ρπ ( b 2 − a 2 )
E=
rˆ
2π r ∈0
*P24.65 (a)
(a < r < b)
(r > b)
Consider a gaussian surface in the shape of a rectangular
box with two faces perpendicular to the direction of the
field. It encloses some charge, so the net flux out of the
box is nonzero. The field must be stronger on one side
than on the other. The field cannot be uniform in magnitude.
+
+
+
+
(b)
Now the volume contains no charge. The net flux out of the
box is zero. The flux entering is equal to the flux exiting.
The field magnitude is uniform at points along one field
line. The field magnitude can vary over the faces of the
box perpendicular to the field.
+
+
+
FIG. P24.65(a)
45
46
Chapter 24
ANSWERS TO EVEN PROBLEMS
P24.2
355 kN ⋅ m 2 C
P24.4
(a) −2.34 kN ⋅ m 2 C
P24.6
(a) −55.7 nC
the shell.
(b) The negative charge has a spherically symmetric distribution concentric with
q
2∈0
charge.
q
2∈0
P24.8
(a)
(b)
(b) +2.34 kN ⋅ m 2 C
(c) 0
(c) Plane and square both subtend a solid angle of a hemisphere at the
(b) 678 kN ⋅ m 2 C
P24.10
(a) 1.36 MN ⋅ m 2 C
P24.12
1.77 pC m 3 positive
P24.14
Q−6 q
6∈0
P24.16
28.2 N ⋅ m 2 C
P24.18
(a) 0
P24.20
(a) 16.2 MN C toward the filament
toward the filament
P24.22
(a) 913 nC
P24.24
(a) A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributed throughout its volume, with density 5.00 m C/m3. Find the magnitude of the electric field it creates at a
point P, 3.00 cm from its axis. As a gaussian surface choose a concentric cylinder with its curved
surface passing through the point P and with length 8.00 cm. (b) 3.77 kN/C
P24.26
4.86 GN C away from the wall. It is constant close to the wall.
P24.28
3.50 kN
P24.30
(a) ~1 mN
P24.32
E = Q/2∈0 A vertically upward in each case if Q > 0
(b) 365 kN C
(c) 1.46 MN C
(c) No; see the solution.
(d) 649 kN C
(b) 8.09 MN C toward the filament
(c) 1.62 MN C
(b) 0
(b) ~100 nC
(c) ~10 kN/C
(d) ~10 kN · m2/C
P24.34
(a) 0 (b) 12.4 kN C radially outward (c) 639 N C radially outward (d) No answer changes.
The solid copper sphere carries charge only on its outer surface.
P24.36
31.9 nC/m3
P24.38
The electric field just outside the surface is given by s /∈0. At this point the uniformly charged
surface of the sphere looks just like a uniform flat sheet of charge.
P24.40
See the solution.
P24.42
chw 2
2
Gauss’s Law
P24.44
(a) 2.00 m C (b) to the right (c) 702 kN/C (d) 0 (e) 0 (f) 3.00 m C (g) 4.21 MN/C
radially outward (h) 1.54 m C (i) 8.63 MN/C radially outward (j) −3.00 m C
(k) 2.00 m C (l) See the solution.
P24.46
See the solution.
P24.48
(a, b) See the solution.
P24.50
(a) 0
P24.52
See the solution.
P24.54
0.269 N ⋅ m 2 C; 2.38 pC
P24.56
See the solution.
P24.58
(a)
P24.60
Cd 3 ˆ
d Cd 3 ˆ
d
i for x > ; E = −
i for x < −
(a) E =
24 ∈0
2
24∈0
2
(b)
ρ0 r
2 ∈0
(c)
σ
to the right
∈0
2r ⎞
⎛
⎜⎝ a − ⎟⎠
3b
(b)
1
2π
ke e 2
me R 3
(d) 102 pm
(c) 0
ρ0 R 2 ⎛
2R⎞
⎟
⎜a −
2 ∈0 r ⎝
3b ⎠
Cx 3
ˆi for x > 0;
(b) E =
3∈0
Cx 3 ˆ
E=−
i for x < 0
3∈0
Qr 3
∈0 a 3
(b)
Q
∈0
P24.62
(a)
(c) See the solution
P24.64
For r < a, E = l/2p ∈0r radially outward.
For a < r < b, E = [l + rp(r2−a2)]/2p ∈0r radially outward.
For r > b, E = [l + rp(b2−a2)]/2p ∈0r radially outward.
47
25
Electric Potential
CHAPTER OUTLINE
25.1
25.2
25.3
25.4
25.5
25.6
25.7
25.8
Electric Potential and Potential
Difference
Potential Difference in a Uniform
Electric Field
Electric Potential and Potential
Energy Due to Point Charges
Obtaining the Value of the
Electric Field from the Electric
Potential
Electric Potential Due to
Continuous Charge Distributions
Electric Potential Due to a
Charged Conductor
The Millikan Oil Drop Experiment
Application of Electrostatics
ANSWERS TO QUESTIONS
Q25.1
When one object B with electric charge is immersed
in the electric field of another charge or charges A, the
system possesses electric potential energy. The energy
can be measured by seeing how much work the field does
on the charge B as it moves to a reference location. We
choose not to visualize A’s effect on B as an action-at-adistance, but as the result of a two-step process: Charge
A creates electric potential throughout the surrounding
space. Then the potential acts on B to inject the system
with energy.
*Q25.2 (i) The particle feels an electric force in the negative x
direction. An outside agent pushes it uphill against this
force, increasing the potential energy. Answer (a).
(ii) The potential decreases in the direction of the electric
field. Answer (c).
*Q25.3
The potential is decreasing toward the bottom of the page, so the electric field is downward.
Answer (f).
*Q25.4
(i) At points off the x axis the electric field has a nonzero y component. At points on the negative
x axis the field is to the right and positive. At points to the right of x = 500 mm the field is to the
left and nonzero. The field is zero at one point between x = 250 mm and x = 500 mm. Answer (b).
(ii) The electric potential is negative at this and at all points. Answer (c). (iii) Answer (d).
(iv) Answer (d).
Q25.5
To move like charges together from an infinite separation, at which the potential energy of the
system of two charges is zero, requires work to be done on the system by an outside agent. Hence
energy is stored, and potential energy is positive. As charges with opposite signs move together
from an infinite separation, energy is released, and the potential energy of the set of charges
becomes negative.
Q25.6
(a) The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge.
(b) The equipotential surfaces are nesting concentric spheres around a uniformly charged sphere.
*Q25.7
Answer (b). The potential could have any value.
*Q25.8
The same charges at the same distance away create the same contribution to the total potential.
Answer (b).
49
50
Chapter 25
*Q25.9 The change in kinetic energy is the negative of the change in electric potential energy, so we
work out −q∆V = −q(Vf − Vi) in each case.
(a) −(−e)(60 V − 40 V) = +20 eV
(b) −(−e)(20 V − 40 V) = −20 eV
(c) −(e)(20 V − 40 V) = +20 eV
(d) −(e)(10 V − 40 V) = +30 eV
(e) −(−2e)(50 V − 40 V) = +20 eV
(f) −(−2e)(60 V − 40 V) = +40 eV
With also (g) 0 and (h) +10 eV, the ranking is f > d > c = e = a > h > g > b.
Q25.10 The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the
air—drier air has a larger dielectric breakdown strength, resulting in a higher attainable electric
potential. If other grounded objects are nearby, the maximum potential might be reduced.
*Q25.11 (i) The two spheres come to the same potential, so q/R is the same for both. With twice the
radius, B has twice the charge. Answer (d).
(ii) All the charge runs away from itself to the outer surface of B. Answer (a).
Q25.12 The grounding wire can be touched equally well to any point on the sphere. Electrons will drain
away into the ground and the sphere will be left positively charged. The ground, wire, and sphere
are all conducting. They together form an equipotential volume at zero volts during the contact. However close the grounding wire is to the negative charge, electrons have no difficulty in
moving within the metal through the grounding wire to ground. The ground can act as an infinite
source or sink of electrons. In this case, it is an electron sink.
SOLUTIONS TO PROBLEMS
Section 25.1
P25.1
(a)
Electric Potential and Potential Difference
Energy of the proton-field system is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V.
K i + U i + ∆Emech = K f + U f
0 + qV + 0 =
1
m v 2p + 0
2
(1.60 × 10
1J ⎞ 1
= 1.67 × 10 −27 kg ) v 2p
C ) (120 V ) ⎛
⎝ 1 V ⋅ C⎠ 2 (
−19
v p = 1.52 × 10 5 m s
(b)
The electron will gain speed in moving the other way,
from Vi = 0 to V f = 120 V:
K i + U i + ∆Emech = K f + U f
0+0+0=
0=
1
m ve2 + qV
2
1
(9.11 × 10 −31 kg) ve2 + ( −1.60 × 10 −19 C) (120 J C)
2
ve = 6.49 × 10 6 m s
P25.2
∆V = −14.0 V
and
Q = − N A e = − ( 6.02 × 10 23 ) (1.60 × 10 −19 ) = −9.63 × 10 4 C
W
,
Q
so
W = Q ∆V = ( −9.63 × 10 4 C ) ( −14.0 J C ) = 1.35 MJ
∆V =
Electric Potential
Section 25.2
51
Potential Difference in a Uniform Electric Field
∆V 25.0 × 10 3 J C
=
= 1.67 × 10 6 N C = 1.67 MN C
d
1.50 × 10 −2 m
P25.3
E=
P25.4
B
C
B VB − VA = − ∫ E ⋅ d s = − ∫ E ⋅ d s − ∫ E ⋅ d s
A
A
C
0.500
VB − VA = ( − E cos 180° )
∫
dy − ( E cos 90.0° )
−0.300
0.400
∫
dx
−0.200
VB − VA = ( 325) ( 0.800 ) = +260 V
FIG. P25.4
P25.5
)
(
)
)
)
2
2
1
1
∆U = − m v 2f − vi2 = − ( 9.11 × 10 −31 kg ⎡(1.40 × 10 5 m s − ( 3.70 × 10 6 m s ⎤
⎣
⎦
2
2
−18
= 6.23 × 10 J
+6.23 × 10 −18 = ( −1.60 × 10 −19 ) ∆V
∆U = q∆V :
∆V = −38.9 V. The origin is at highest potentiial.
P25.6
Assume the opposite. Then at some point A on some equipotential surface the electric field has a
nonzero component E p in the plane of the surface. Let a test charge start from point A and move
B
some distance on the surface in the direction of the field component. Then ∆V = − ∫ E ⋅ d s is
A
nonzero. The electric potential charges across the surface and it is not an equipotential surface. The
contradiction shows that our assumption is false, that E p = 0, and that the field is perpendicular to the
equipotential surface.
P25.7
(a)
Arbitrarily choose V = 0 at 0. Then at other points
V = − Ex
and
U e = QV = −QEx
Between the endpoints of the motion,
( K + U s + U e )i = ( K + U s + U e ) f
1 2
0 + 0 + 0 = 0 + kxmax
− QExmax
2
(b)
FIG. P25.7
At equilibrium,
∑F
x
= − Fs + Fe = 0
or
kx = QE
So the equilibrium position is at x =
(c)
xmax
so
2QE
=
k
QE
.
k
The block’s equation of motion is
∑F
QE
,
k
QE
,
k
Let
x′ = x −
or
x = x′ +
x
= − kx + QE = m
d2x
.
dt 2
continued on next page
52
Chapter 25
so the equation of motion becomes:
d 2 ( x + QE k )
QE ⎞
+ QE = m
−k ⎛ x′ +
,
⎝
k ⎠
dt 2
or
d2x′
k
= − ⎛ ⎞ x′
⎝ m⎠
dt 2
This is the equation for simple harmonic motion ax′ = −ω 2 x ′
(d)
with
ω=
The period of the motion is then
T=
k
m
2π
m
= 2π
ω
k
( K + U s + U e )i + ∆Emech = ( K + U s + U e ) f
1 2
0 + 0 + 0 − µ k mgxmax = 0 + kxmax
− QExmax
2
xmax =
P25.8
2 (QE − µ k mg )
k
Arbitrarily
take V = 0 at point P. Then the potential at the original position of the charge is
− E ⋅ s = − EL cos θ . At the final point a, V = −EL. Because the table is frictionless we have
( K + U )i = ( K + U ) f
0 − qEL cos θ =
v=
P25.9
2qEL (1 − cos θ )
=
m
2 ( 2.00 × 10 −6 C ) ( 300 N C ) (1.50 m ) (1 − cos 60.0° )
0.010 0 kg
= 0.300 m s
Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length,
V = −Ed and U e = − λ LEd .
(a)
( K + U )i = ( K + U ) f
0+0=
v=
(b)
Section 25.3
P25.10
1
m v 2 − qEL
2
1
µ L v 2 − λ LEd
2
2λ Ed
=
µ
2 ( 40.0 × 10 −6 C m ) (100 N C ) ( 2.00 m )
( 0.100
kg m )
= 0.400 m s
The same. Each bit of the rod feels a force of the same size as before.
Electric Potential and Potential Energy Due to Point Charges
(a)
Since the charges are equal and placed symmetrically, F = 0 .
(b)
Since F = qE = 0, E = 0 .
(c)
V = 2 ke
⎛ 2.00 × 10 −6 C ⎞
q
= 2 (8.99 × 10 9 N ⋅ m 2 C2 ) ⎜
⎝ 0.800 m ⎟⎠
r
V = 4.50 × 10 4 V = 45.0 kV
FIG. P25.10
Electric Potential
P25.11
(a)
The potential at 1.00 cm is
V1 = ke
2
2
−19
9
q (8.99 × 10 N ⋅ m C ) (1.60 × 10 C )
= 1.44 × 10 −7 V
=
−2
r
1.00 × 10 m
(b) The potential at 2.00 cm is
V2 = ke
2
2
−19
9
q (8.99 × 10 N ⋅ m C ) (1.60 × 10 C )
= 0.719 × 10 −7 V
=
−2
2.00 × 10 m
r
Thus, the difference in potential between the two points is ∆V = V2 − V1 = −7.19 × 10 −8 V .
(c)
The approach is the same as above except the charge is −1.60 × 10 −19 C. This changes the
sign of each answer, with its magnitude remaining the same.
That is, the potential at 1.00 cm is −1.44 × 10 −7 V .
The potential at 2.00 cm is −0.719 × 10 −7 V,
P25.12
(a)
Ex =
ke q1
ke q2
+
2 = 0
x 2 ( x − 2.00 )
so
⎛ +q
−2q ⎞
E x = ke ⎜ 2 +
=0
( x − 2.00 )2 ⎟⎠
⎝x
becomes
2qx 2 = q ( x − 2.00 )
Dividing by ke ,
∆V = V2 − V1 = 7.19 × 10 −8 V
2
x 2 + 4.00 x − 4.00 = 0
−4.00 ± 16.0 + 16.0
= −4.83 m
2
(Note that the positive root does not correspond to a physically valid situation.)
Therefore E = 0
(b)
V=
ke q1
k e q2
+
=0
2.00 − x
x
when
or
and
P25.13
(a)
(b)
q
−2q
=
x
2− x
when
x = 0.667 m
For x < 0
x = −2.00 m
E=
Q
4π ∈0 r 2
V=
Q
4π ∈0 r
r=
V
3 000 V
=
= 6.00 m
E 500 V m
V = −3 000 V =
Q=
+q
2q ⎞
V = ke ⎛
−
=0
⎝ x 2.00 − x ⎠
2qx = q ( 2.00 − x )
Again solving for x,
For 0 ≤ x ≤ 2.00 V = 0
x=
Q
4π ∈0 ( 6.00 m )
−3 000 V
( 6.00 m ) = −2.00 µ C
(8.99 × 109 V ⋅ m C)
53
54
P25.14
Chapter 25
U=
(a)
(5.00 × 10
qQ
=
4π ∈0 r
−9
C ) ( −3.00 × 10 −9 C ) (8.99 × 10 9 V ⋅ m C )
( 0.350 m )
= −3.86 × 10 −7 J
The minus sign means it takes 3.86 × 10 −7 J to pull the two charges apart from 35 cm to a
much larger separation.
V=
(b)
=
Q1
Q2
+
4π ∈0 r1 4π ∈0 r2
(5.00 × 10
−9
C ) (8.99 × 10 9 V ⋅ m C )
0.175 m
+
( −3.00 × 10
−9
C ) (8.99 × 10 9 V ⋅ m C )
0.175 m
V = 103 V
P25.15
V = ∑k
i
qi
ri
1 ⎤
1
⎡ −1
−
+
V = (8.99 × 10 9 ) ( 7.00 × 10 −6 ) ⎢
⎥
⎣ 0.010 0 0.010 0 0.038 7 ⎦
V = −1.10 × 10 7 V = −11.0 MV
FIG. P25.15
P25.16
(a)
V=
ke q1 ke q2
kq
+
= 2⎛ e ⎞
⎝ r ⎠
r1
r2
⎛ (8.99 × 10 9 N ⋅ m 2 C2 ) ( 2.00 × 10 −6 C ) ⎞
V = 2⎜
⎟
(1.00 m )2 + ( 0.500 m )2
⎠
⎝
V = 3.22 × 10 4 V = 32.2 kV
(b)
P25.17
FIG. P25.16
U = qV = ( −3.00 × 10 −6 C ) ( 3.22 × 10 4 J C ) = −9.65 × 10 −2 J
⎛ 1 ⎞ ⎛ q1 q2 q3 ⎞
U e = q4V1 + q4V2 + q4V3 = q4 ⎜
+ +
⎝ 4π ∈0 ⎟⎠ ⎜⎝ r1 r2 r3 ⎟⎠
⎛
⎞
2
1
1
1
U e = (10.0 × 10 −6 C ) (8.99 × 10 9 N ⋅ m 2 C2 ) ⎜
+
+
⎟
( 0.600 m )2 + ( 0.150 m )2 ⎠
⎝ 0.600 m 0.150 m
U e = 8.95 J
Electric Potential
*P25.18 (a)
P25.19
55
The first expression, with distances squared, describes an electric field. The second expression
describes an electric potential. Then a positive 7 nC charge is 7 cm from the origin. To create
field that is to the left and downward, it must be in the first quadrant, with position
vector 7 cm at 70° . A negative 8 nC charge 3 cm from the origin creates an upward
(b)
electric field at the origin, so it must be at 3 cm at 90° . We evaluate the given expressions:
E = −4.39 kN C ˆi + 67.8 kN C ˆj
V = −1.50 kV
F = qE = −16 × 10 −9 C −4.39 ˆi + 67.8 ˆj 10 3 N C = 7.03ˆi − 109 ˆj × 10 −5 N
(c)
U e = qV = −16 × 10 −9 C ( −1.50 × 10 3 J C ) = +2.40 × 10 −5 J
(
)
(
)
U = U1 + U 2 + U 3 + U 4
U = 0 + U12 + (U13 + U 23 ) + (U14 + U 24 + U 34 )
U = 0+
U=
2
keQ 2 keQ 2 ⎛ 1
1
⎞ kQ ⎛
⎞
+
+ 1⎟ + e ⎜ 1 +
+ 1⎟
⎜⎝
⎠
⎝
s
s
s
2
2 ⎠
keQ 2 ⎛
keQ 2
2 ⎞
⎜⎝ 4 +
⎟⎠ = 5.41
s
s
2
FIG. P25.19
2 ⎞
⎛
We can visualize the term ⎜ 4 +
⎟ as arising directly from the 4 side pairs and 2 face diagonal
⎝
2⎠
pairs.
P25.20
Each charge creates equal potential at the center. The total potential is:
5k q
⎡ k (−q) ⎤
V = 5⎢ e
= − e
⎥
R
⎣ R ⎦
P25.21
(a)
Each charge separately creates positive potential everywhere. The total potential produced
by the three charges together is then the sum of three positive terms. There is no point ,
at a finite distance from the charges, at which this total potential is zero.
(b)
V=
ke q ke q
2 ke q
+
=
a
a
a
56
P25.22
Chapter 25
(a)
V ( x) =
k ( +Q )
ke ( +Q )
keQ1 keQ2
+
+
= e2
2
2
r1
r2
x +a
x2 + (−a)
2 keQ
V ( x) =
V ( x)
x 2 + a2
( keQ a )
(b)
V ( y) =
=
=
keQ ⎛
⎜
a ⎜⎝
⎞
⎟
( x a )2 + 1 ⎟⎠
2
2
( x a )2 + 1
keQ1 keQ2 ke ( +Q ) ke ( −Q )
+
=
+
r1
r2
y−a
y+a
FIG. P25.22(a)
k Q⎛ 1
1 ⎞
V ( y) = e ⎜
−
a ⎝ y a − 1 y a + 1 ⎟⎠
⎛ 1
V ( y)
1 ⎞
= ⎜
−
( keQ a ) ⎝ y a − 1 y a + 1 ⎟⎠
FIG. P25.22(b)
P25.23
Consider the two spheres as a system.
(a)
( )
v2 =
Conservation of momentum:
0 = m1 v1 ˆi + m2 v2 − ˆi
By conservation of energy,
0=
and
ke q1q2 ke q1q2 1
1 m12 v12
−
= m1 v12 +
r1 + r2
d
2
2 m2
or
m1 v1
m2
ke ( − q1 ) q2 1
k ( − q1 ) q2
1
= m1 v12 + m2 v22 + e
d
r1 + r2
2
2
v1 =
2 m2 ke q1q2 ⎛ 1
1⎞
− ⎟
⎜
m1 ( m1 + m2 ) ⎝ r1 + r2 d ⎠
v1 =
2 ( 0.700 kg ) (8.99 × 10 9 N ⋅ m 2 C2 ) ( 2 × 10 −6 C ) ( 3 × 10 −6 C ) ⎛
1
1 ⎞
−
−3
⎝
8 × 10 m 1.00 m ⎠
( 0.100 kg) ( 0.800 kg)
= 10.8 m s
v2 =
(b)
m1 v1 0.100 kg (10.8 m s )
=
= 1.55 m s
0.700 kg
m2
If the spheres are metal, electrons will move around on them with negligible energy loss to
place the centers of excess charge on the insides of the spheres. Then just before they touch,
the effective distance between charges will be less than r1 + r2 and the spheres will really
be moving faster than calculated in (a) .
Electric Potential
P25.24
Consider the two spheres as a system.
(a)
( )
Conservation of momentum:
0 = m1 v1 ˆi + m2 v2 − ˆi
or
v2 =
By conservation of energy,
0=
and
ke q1q2 ke q1q2 1
1 m12 v12
.
−
= m1 v12 +
r1 + r2
d
2
2 m2
v1 =
m1 v1
.
m2
ke ( − q1 ) q2 1
k ( − q1 ) q2
1
= m1 v12 + m2 v22 + e
d
r1 + r2
2
2
2 m2 ke q1q2 ⎛ 1
1⎞
− ⎟
⎜
m1 ( m1 + m2 ) ⎝ r1 + r2 d ⎠
⎛m ⎞
v2 = ⎜ 1 ⎟ v1 =
⎝ m2 ⎠
(b)
P25.25
57
2 m1 ke q1q2 ⎛ 1
1⎞
− ⎟
⎜
m2 ( m1 + m2 ) ⎝ r1 + r2 d ⎠
If the spheres are metal, electrons will move around on them with negligible energy loss to
place the centers of excess charge on the insides of the spheres. Then just before they touch,
the effective distance between charges will be less than r1 + r2 and the spheres will really
be moving faster than calculated in (a) .
The original electrical potential energy is
U e = qV = q
ke q
d
In the final configuration we have mechanical equilibrium. The spring and electrostatic forces
kq
k q2
on each charge are − k ( 2d ) + q e 2 = 0. Then k = e 3 . In the final configuration the total
18d
( 3d )
1 2
1 ke q 2
ke q 4 ke q 2
2
=
. The missing energy must have
potential energy is kx + qV =
( 2d ) + q
2
2 18d 3
3d 2 9 d 2
kq
4k q
become internal energy, as the system is isolated: e = e + ∆ Eint .
d
9d
∆Eint =
P25.26
5 ke q 2
9 d
Using conservation of energy for the alpha particle-nucleus system,
we have
K f + U f = K i + Ui
But
Ui =
ke qα qgold
Also
and ri ≈ ∞ Thus,
ri
K f = 0 (v f = 0 at turning point),
so
U f = Ki
or
ke qα qgold
rmin
rmin =
=
Ui = 0
1
mα vα2
2
2 ke qα qgold
mα vα2
= 27.4 fm
=
2 (8.99 × 10 9 N ⋅ m 2 C2 ) ( 2 ) ( 79 ) (1.60 × 10 −19 C )
(6.64 × 10 −27 kg) ( 2.00 × 10 7 m s )
2
2
= 2.74 × 10 −14 m
58
P25.27
Chapter 25
Each charge moves off on its diagonal line. All charges have equal speeds.
∑ (K + U ) = ∑ (K + U )
i
2
0+
f
2
4 ke q
2k q
1
4 k q 2 2k q 2
+ e = 4 ⎛ mv 2 ⎞ + e + e
⎝2
⎠
2L
L
2L
2 2L
1 ⎞ ke q 2
⎛
= 2mv 2
⎜⎝ 2 +
⎟
2⎠ L
1 ⎞ ke q 2
⎛
v = ⎜1 +
⎟
⎝
8 ⎠ mL
P25.28
A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12
face diagonal pairs separated by 2s, and 4 interior diagonal pairs separated by 3s.
U=
ke q 2
s
Section 25.4
P25.29
P25.31
Obtaining the Value of the Electric Field from the Electric Potential
V = a + bx = 10.0 V + ( −7.00 V m ) x
(a)
P25.30
12
4 ⎤
ke q 2
⎡
⎢⎣12 + 2 + 3 ⎥⎦ = 22.8 s
At x = 0,
V = 10.0 V
At x = 3.00 m,
V = −11.0 V
At x = 6.00 m,
V = −32.0 V
dV
= −b = − ( −7.00 V m ) = 7.00 N C in the + x direection
dx
(b)
E=−
(a)
For r < R
(b)
For r ≥ R
keQ
R
dV
Er = −
= 0
dr
kQ
V= e
r
dV
kQ
kQ
Er = −
= − ⎛ − e2 ⎞ = e2
⎝
⎠
dr
r
r
V=
V = 5 x − 3 x 2 y + 2 yz 2
Evaluate E at (1, 0, − 2 ) .
∂V
= −5 + 6 xy = −5 + 6 (1) ( 0 ) = −5
∂x
∂V
2
2
Ey = −
= +3 x 2 − 2 z 2 = 3 (1) − 2 ( −2 ) = −5
∂y
Ex = −
Ez = −
∂V
= −4 yz = −4 ( 0 ) ( −2 ) = 0
∂z
E = Ex2 + Ey2 + Ez2 = ( −5) + ( −5) + 0 2 = 7.07 N C
2
2
Electric Potential
P25.32
P25.33
EA > EB since E =
(a)
∆V
∆s
∆V
(6 − 2) V
=−
= 200 N C down
∆s
2 cm
(b)
EB = −
(c)
The figure is shown to the right, with sample field lines
sketched in.
Ey = −
Ey =
Section 25.5
∂V
∂ ⎡ k Q ⎛ + 2 + y2 ⎞ ⎤
= − ⎢ e ln ⎜
⎟⎥
∂y
∂y ⎢ y
⎝
⎠ ⎥⎦
⎣
keQ ⎡
y2
⎢1 − 2
y ⎢⎣ + y 2 + 2 + y 2
⎤
keQ
⎥=
y 2 + y2
⎥⎦
Electric Potential Due to Continuous Charge Distributions
keQ
∆V = V2 R − V0 =
P25.35
(a)
λ
C 1
C
[α ] = ⎡⎢ ⎤⎥ = ⋅ ⎛⎝ ⎞⎠ = 2
m
⎣x⎦ m m
(b)
V = ke ∫
V=∫
α xdx
ke dq
= ke ∫
2
2
r
b + ( L 2 − x)
Let z =
R + (2R)
2
2
−
keQ keQ ⎛ 1
kQ
⎞
=
− 1⎟ = −0.553 e
⎜⎝
⎠
R
R
R
5
P25.34
P25.36
FIG. P25.32
λ dx
dq
xdx
L ⎤
⎡
= ke ∫
= keα ∫
= keα ⎢ L − d ln ⎛ 1 + ⎞ ⎥
⎝
r
r
d
+
x
d⎠⎦
⎣
0
L
FIG. P25.35
L
− x.
2
Then x =
L
− z, and dx = − dz
2
V = keα ∫
(L
2 − z ) ( − dz )
=−
keα L
dz
zdz
+ keα ∫ 2 2
2 ∫ b2 + z2
b +z
b +z
keα L
ln z + z 2 + b 2 + keα z 2 + b 2
=−
2
2
(
2
)
2
⎤
k α L ⎡⎛ L
L
V =− e
ln ⎢
− x⎞ + ⎛ − x⎞ + b 2 ⎥
⎠
⎝2
⎠
2
⎢⎣⎝ 2
⎥⎦
2
keα L ⎡ L 2 − L + ( L 2 ) + b
ln ⎢
2
⎢ L 2 + ( L 2 )2 + b 2
⎣
2
V =−
2
2
keα L ⎡⎢ b + ( L 4 ) − L 2 ⎤⎥
V= −
ln
2
⎢ b 2 + ( L2 4 ) + L 2 ⎥
⎣
⎦
L
2
L
+ keα ⎛ − x ⎞ + b 2
⎝2
⎠
0
L
0
2
2
⎤
⎡ L
⎤
L
⎥ + keα ⎢ ⎛ − L ⎞ + b 2 − ⎛ ⎞ + b 2 ⎥
⎠
⎝ 2⎠
⎥
⎢⎣ ⎝ 2
⎦⎥
⎦
59
60
P25.37
Chapter 25
V = ∫ dV =
1
4π ∈0
∫
dq
r
All bits of charge are at the same distance from O.
So V =
⎛ −7.50 × 10 −6 C ⎞
1 ⎛ Q⎞
= (8.99 × 10 9 N ⋅ m 2 C2 ) ⎜
= −1.51 MV .
⎝ 0.140 m π ⎟⎠
4π ∈0 ⎝ R ⎠
−R
P25.38
V = ke
dq
λ dx
λ ds
λ dx
= ke ∫
+ ke ∫
+ ke ∫
∫
r
−
x
R
x
R
all charge
−3 R
semicircle
−R
V = − ke λ ln ( − x ) −3 R +
V = ke ln
Section 25.6
P25.39
(a)
3R
ke λ
3R
π R + ke λ ln x R
R
3R
+ ke λ π + ke ln 3 = ke λ (π + 2 ln 3)
R
Electric Potential Due to a Charged Conductor
E= 0 ;
9
−6
ke q (8.99 × 10 ) ( 26.0 × 10 )
V=
=
= 1.67 MV
R
0.140
(b)
(c)
P25.40
E=
9
−6
ke q (8.99 × 10 ) ( 26.0 × 10 )
=
= 5.84 MN C away
r2
( 0.200 )2
V=
9
−6
ke q (8.99 × 10 ) ( 26.0 × 10 )
=
= 1.17 MV
R
0.200
E=
9
−6
ke q (8.99 × 10 ) ( 26.0 × 10 )
=
= 11.9 MN C away
R2
( 0.140 )2
V=
ke q
= 1.67 MV
R
Substituting given values into V =
ke q
r
7.50 × 10 3 V =
(8.99 × 10
9
N ⋅ m 2 C2 ) q
0.300 m
−7
Substituting q = 2.50 × 10 C,
N=
2.50 × 10 −7 C
= 1.56 × 1012 electrrons
1.60 × 10 −19 C e−
Electric Potential
P25.41
The electric field on the surface of a conductor varies inversely with the radius of curvature of the
surface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa.
The local charge density and the electric field intensity are related by
E=
(a)
σ
∈0
or
σ = ∈0 E
Where the radius of curvature is the greatest,
σ = ∈0 Emin = (8.85 × 10 −12 C2 N ⋅ m 2 ) ( 2.80 × 10 4 N C ) = 248 nC m 2
(b)
Where the radius of curvature is the smallest,
σ = ∈0 Emax = (8.85 × 10 −12 C2 N ⋅ m 2 ) ( 5.60 × 10 4 N C ) = 496 nC m 2
P25.42
61
(a)
Both spheres must be at the same potential according to
where also
q1 + q2 = 1.20 × 10 −6 C
Then
q1 =
ke q1 ke q2
=
r1
r2
q2 r1
r2
q2 r1
+ q2 = 1.20 × 10 −6 C
r2
q2 =
1.20 × 10 −6 C
= 0.300 × 10 −6 C on the smaller sphere
1 + 6 cm 2 cm
q1 = 1.20 × 10 −6 C − 0.300 × 10 −6 C = 0.900 × 10 −6 C
V=
(b)
2
2
−6
9
ke q1 (8.99 × 10 N ⋅ m C ) ( 0.900 × 10 C )
=
= 1.35 × 10 5 V
r1
6 × 10 −2 m
Outside the larger sphere,
kq
V
1.35 × 10 5 V
E1 = e 2 1 rˆ = 1 rˆ =
rˆ = 2.25 × 10 6 V m away
r1
r1
0.06 m
Outside the smaller sphere,
1.35 × 10 5 V
E2 =
rˆ = 6.74 × 10 6 V m away
0.02 m
The smaller sphere carries less charge but creates a much stronger electric field than the
larger sphere.
62
Chapter 25
Section 25.7
The Millikan Oil Drop Experiment
Section 25.8
Application of Electrostatics
P25.43
(a)
Emax = 3.00 × 10 6 V m =
1
keQ keQ ⎛ 1 ⎞
= Vmax ⎛ ⎞
=
2
⎝
⎝
⎠
r
r r
r⎠
Vmax = Emax r = 3.00 × 10 6 ( 0.150 ) = 450 kV
{
keQmax
= Vmax
r
}
Emax r 2 3.00 × 10 6 ( 0.150 )2
=
= 7.51 µ C
8.99 × 10 9
ke
(b)
keQmax
= Emax
r2
(a)
VB − VA = − ∫ E ⋅ ds and the field at distance r from a uniformly
or
Qmax =
B
P25.44
A
charged rod (where r > radius of charged rod) is
E=
λ
2k λ
= e
r
2π ∈0 r
In this case, the field between the central wire and the coaxial
cylinder is directed perpendicular to the line of charge so that
rb
VB − VA = − ∫
ra
or
(b)
FIG. P25.44
⎛r ⎞
2 ke λ
dr = 2 ke λ ln ⎜ a ⎟
r
⎝ rb ⎠
⎛r ⎞
∆V = 2 ke λ ln ⎜ a ⎟
⎝ rb ⎠
From part (a), when the outer cylinder is considered to be at zero potential, the potential at
a distance r from the axis is
r
V = 2 ke λ ln ⎛ a ⎞
⎝ r⎠
The field at r is given by
E=−
⎛ r⎞
2k λ
∂V
r
= −2 ke λ ⎜ ⎟ ⎛ − a2 ⎞ = e
⎝
⎠
∂r
r
r
⎝ ra ⎠
But, from part (a), 2ke λ =
Therefore, E =
∆V
.
ln ( ra rb )
∆V ⎛ 1 ⎞
.
ln ( ra rb ) ⎝ r ⎠
Electric Potential
P25.45
(a)
63
From the previous problem,
E=
∆V 1
ln ( ra rb ) r
We require just outside the central wire
5.50 × 10 6 V m =
or
50.0 × 10 3 V ⎛ 1 ⎞
ln ( 0.850 m rb ) ⎜⎝ rb ⎠⎟
⎛
⎞
m
(110 m ) r ln ⎜⎝ 0.850
⎟⎠ = 1
r
−1
b
b
We solve by homing in on the required value
rb ( m )
⎛
⎞
m
(110 m ) r ln ⎜⎝ 0.850
⎟⎠
r
−1
b
0.0100
0.00100
0.00150
0.00145
4.89
0.740
1.05
1.017
b
Thus, to three significant figures,
rb = 1.42 mm
(b)
At ra ,
E=
50.0 kV
1
⎛
⎞ = 9.20 kV m
ln ( 0.850 m 0.001 42 m ) ⎝ 0.850 m ⎠
0.00143 0.00142
1.005
0.999
64
Chapter 25
Additional Problems
*P25.46 (a)
(b)
The two particles exert forces of repulsion on each other. As the projectile approaches the
target particle, the projectile slows. The target starts to move in the x direction. As long as
the projectile is moving faster than the second particle, the two will be approaching. Kinetic
energy will be being converted into electric potential energy. When both particles move
with equal speeds, the distance between them will momentarily not be changing: this is
the instant of closest approach. Thereafter, the target particle, still feeling a forward force,
will move faster than the projectile. The particles will separate again. The particles exert
forces on each other but never touch. The particles will eventually be very far apart, with
zero electric potential energy. All of the Ue they had at closest approach is converted back
into kinetic energy. The whole process is an elastic collision. Compare this problem with
Problem 9.49 in Chapter 9.
Momentum is constant throughout the process. We equate it at the large-separation initial
point and the point b of closest approach.
m1v1i + m2 v 2i = m1v1b + m2 v 2b
( 2 g) 21ˆi m s + 0 = ( 2 g + 5 g) vb
)
(
v b = 6.00 ˆi m s
(c)
Energy conservation between the same two points:
1
1
1
kqq
m1 v12i + m2 v22i + 0 = ( m1 + m2 ) vb2 + e 1 2
2
2
2
rb
1
1
1
2
2
2
0.002 kg ( 21 m s ) + 0.005 kg ( 0 ) + 0 = 0.007 kg ( 6 m s )
2
2
2
8.99 × 10 9 Nm 2 15 × 10 −6 C 8.5 × 10 −6 C
+
C2
rb
1.15 J ⋅ m
0.441 J − 0.126 J =
rb
1.15 m
rb =
= 3.64 m
0.315
(d)
The overall elastic collision is described by conservation of momentum:
m1v1i + m2 v2 i = m1v1d + m2 v2 d
( 2 g) 21ˆi m s + 0 = 2 gv1d ˆi + 5 gv2 d ˆi
)
(
and by the relative velocity equation:
v1i − v2 i = v2 d − v1d
21 m s − 0 = v2 d − v1d
we substitute
v2 d = 21 m s + v1d
42 g ⋅ m s = 2 gv1d + 5 g ( 21 m s + v1d ) = 2 gv1d + 105 g ⋅ m s + 5 gv1d
−63 g ⋅ m s = 7 gv1d
v1d = −99.00 m s
v1d = −9.00 ˆi m s
v2 d = 21 m s − 9 m s = 12.0 m s
v 2 d = 12.0 ˆi m s
Electric Potential
65
( 38 ) ( 54 ) (1.60 × 10 )
q1q2
= (8.99 × 10 9 )
= 4.04 × 10 −11 J = 253 MeV
r12
( 5.50 + 6.20 ) × 10 −15
−19 2
P25.47
U = qV = ke
*P25.48 (a)
The field within the conducting Earth is zero. E = s /¨0
s = E¨0 = (−120 N/C)(8.85 × 10−12 C2/N⋅m2) = −1.06 nC/m2
(b)
QE = s A = s 4p r2 = (−1.06 × 10−9 C/m2) 4p (6.37 × 106 m)2 = −542 kC
(c)
V=
(d)
Vhead − Vfeet = Ed = (120 N/C)1.75 m = 210 V
(e)
F=
(f)
The gravitational force is
keQ 8.99 × 10 9 C2 (−5.42 × 10 5 C)
=
= −764 MV
R
N ⋅ m 2 (6.37 × 10 6 m)
ke q1q2 8.99 × 10 9 N ⋅ m 2 (5.42 × 10 5 C)2 (0.273)
=
= 4.88 × 10 3 N away from Earth
r2
C2 (3.84 × 108 m)2
GME MM 6.67 × 10 −11 N ⋅ m 2 (5.98 × 10 24 kg)(7.36 × 10 22 kg)
=
= 1.99 × 10 20 N
kg 2 ( 3.84 × 10 8 m)2
r2
toward the Earth
F=
The gravitational force is larger by 1.99 × 1020/4.88 × 103 = 4.08 × 1016 times and in the
opposite direction.
Electrical forces are negligible in accounting for planetary motion.
(g)
We require m(−g) + qE = 0
6 × 10−6 kg(−9.8 m/s2) + q(−120 N/C) = 0
q = 5.88 × 10−5 N/(−120 N/C) = −490 nC
(h)
Less charge to be suspended at the equator. The gravitational force is weaker at a greater
distance from the Earth’s center. The suspended particle is not quite in equilibrium, but
accelerating downward to participate in the daily rotation. At uniform potential, the planet’s
surface creates a stronger electric field at the equator, where its radius of curvature is smaller.
(a)
U=
9
−19
ke q1q2 − ( 8.99 × 10 ) (1.60 × 10 )
=
= −4.35 × 10 −18 J = −27.2 eV
r
0.052 9 × 10 −9
(b)
U=
9
−19
ke q1q2 − (8.99 × 10 ) (1.60 × 10 )
=
= −6.80 eV
r
22 ( 0.052 9 × 10 −9 )
2
P25.49
2
(c)
P25.50
(a)
ke q1q2 − ke e2
=
= 0
r
∞
To make a spark 5 mm long in dry air between flat metal plates requires potential difference
U=
)
)
∆V = Ed = ( 3 × 10 6 V m ( 5 × 10 −3 m = 1.5 × 10 4 V ~10 4 V
(b)
2
The area of your skin is perhaps 1.5 m , so model your body as a sphere with this surface
area. Its radius is given by 1.5 m 2 = 4π r 2 , r = 0.35 m. We require that you are at the
potential found in part (a):
V=
ke q
r
q=
Vr 1.5 × 10 4 V ( 0.35 m ) ⎛ J ⎞ ⎛ N ⋅ m ⎞
=
ke 8.99 × 10 9 N ⋅ m 2 C2 ⎝ V ⋅ C ⎠ ⎝ J ⎠
q = 5.8 × 10 −7 C ~10 −6 C
66
Chapter 25
*P25.51 (a) We have keQ/R = 200 V and keQ/(R + 10 cm) = 150 V
Then 200 R = 150(R + 10 cm). The information is sufficient to determine the charge and
original distance, according to 50 R = 1 500 cm.
R = 30.0 cm
and Q = 200 V(0.3 m)/(8.9 × 109 N⋅m2/C2) = 6.67 nC
(b) We have keQ/R = 210 V and keQ/(R + 10 cm)2 = 400 V/m.
Then 210 VR = (400 V/m)(R + 0.1 m)2.
210 mR = 400 R2 + 80 mR = 4 m2
The information is not quite sufficient . There are two possibilities, according to
400 R 2 − 130 mR + 4 m2 = 0
R=
+130 ± 130 2 − 4(400)4
m = either 0.291 m or 0.034 4 m
800
If the radius is 29.1 cm, the charge is 210(0.291) C/8.99 × 109 = 6.79 nC. If the radius is
3.44 cm, the charge is 210(0.034 4) C/8.99 × 109 = 804 pC.
P25.52
The plates create uniform electric field to the right in the picture, with magnitude
V0 − ( −V0 ) 2V0
. Assume the ball swings a small distance x to the right. It moves to a
=
d
d
2V
place where the voltage created by the plates is lower by − Ex = − 0 x . Its ground
d
connection maintains it at V = 0 by allowing charge q to flow from ground onto the ball,
4V 2 xR
2V x k q
2V xR
where − 0 + e = 0
q = 0 . Then the ball feels electric force F = qE = 0 2 to
d
R
ke d
ke d
the right. For equilibrium this must be balanced by the horizontal component of string
4V 2 xR
4V02 xR x
tension according to T cos θ = mg
for small x.
T sin θ = 0 2 tan θ =
=
ke d 2 mg L
ke d
1
2
⎛ k d 2 mg ⎞
. If V0 is less than this value, the only equilibrium position of the ball is
Then V0 = ⎜ e
⎝ 4 RL ⎟⎠
hanging straight down. If V0 exceeds this value the ball will swing over to one plate or the other.
*P25.53 For a charge at (x = −1 m, y = 0), the radial distance away is given by
term will be the potential it creates if
(8.99 × 109 N ⋅ m2/C2)Q1 = 36 V ⋅ m
Q1 = 4.00 nC
The second term is the potential of a charge at (x = 0, y = 2 m) with
(8.99 × 109 N ⋅ m2/C2)Q2 = −45 V ⋅ m
Q2 = −5.01 nC
Thus we have 4.00 nC at (−1.00 m, 0) and −5.01 nC at (0, 2.00 m) .
( x + 1)2 + y 2 . So the first
Electric Potential
P25.54
67
(a) Take the origin at the point where we will find the potential. One ring, of width dx, has
Qdx
charge
and, according to Example 25.5, creates potential
h
keQdx
dV =
h x 2 + R2
The whole stack of rings creates potential
∫
V=
d +h
∫
dV =
all charge
h x +R
2
d
2
=
(
keQ
ln x + x 2 + R 2
h
)
d +h
d
2
keQ ⎛ d + h + ( d + h ) + R 2 ⎞
ln ⎜
⎟
h
d + d 2 + R2
⎝
⎠
=
(b)
keQdx
Qdx
Qdx
A disk of thickness dx has charge
and charge-per-area
. According to
π R2 h
Example 25.6, it creates potential h
dV = 2π ke
Qdx
π R2 h
(
x 2 + R2 − x
)
Integrating,
d +h
V=
∫
d
2 keQ
R2 h
(
)
x 2 + R 2 dx − xdx =
⎡
keQ ⎢
(d + h)
R2 h ⎢
⎣
V=
Q
P25.55
W = ∫ Vdq
where V =
0
*P25.56 (a)
⎛ d + h + ( d + h )2 + R 2 ⎞ ⎤
⎟⎥
d 2 + R 2 − 2 dh − h 2 + R 2 ln ⎜
⎜⎝
⎟⎠ ⎥
d + d 2 + R2
⎦
Therefore, W =
keQ 2
.
2R
kq
ke q
kq
kq
kq
k q 2k q
− e =+ e − e = e − e = − e
4a
3a + a 3a − a 4 a
4aa
r+a r−a
The approximate expression −2keqa/x2 gives −2keqa/(3a)2 = −keq/4.5a.
Even though 3a is hardly a large distance compared to the 2a separation of the charges,
the approximate answer is different by only
1 / 4 − 1 / 4.5 0.5
=
= 11.1%
1/ 4
4.5
2
2
V2 − V1 = − ∫ E ⋅ dr = − ∫
r
P25.57
)
The exact potential is
+
(b)
( d + h )2 + R 2 − d
ke q
.
R
(
d +h
2 keQ ⎡ 1
R2
x2 ⎤
2
2
2
2
ln
x
x
R
x
x
+
R
+
+
+
−
2
2 ⎥⎦ d
R 2 h ⎢⎣ 2
r1
V2 − V1 =
r
λ
dr
2π ∈0 r
r1
⎛r ⎞
−λ
ln ⎜ 2 ⎟
2π ∈0 ⎝ r1 ⎠
68
P25.58
Chapter 25
Take the illustration presented with the problem as an initial picture.
No external horizontal forces act on the set of four balls, so its center
of mass stays fixed at the location of the center of the square. As the
charged balls 1 and 2 swing out and away from each other, balls 3
and 4 move up with equal y-components of velocity. The maximumkinetic-energy point is illustrated. System energy is conserved:
v
1
+
P25.59
v=
CM
3
2
+
4
v
v
FIG. P25.58
ke q 2 ke q 2 1
1
1
1
=
+ mv 2 + mv 2 + mv 2 + mv 2
a
3a 2
2
2
2
2 ke q 2
= 2 mv 2
3a
v
ke q 2
3am
From an Example in the chapter text, the potential at the center of the
kQ
ring is Vi = e and the potential at an infinite distance from the
R
ring is V f = 0. Thus, the initial and final potential energies of the
point charge-ring system are:
U i = QVi =
keQ 2
R
FIG. P25.59
U f = QV f = 0
and
From conservation of energy,
K f + U f = K i + Ui
or
1
k Q2
M v 2f + 0 = 0 + e
2
R
giving
vf =
a+ L
P25.60
λ dx
(a)
(b)
(x
∫
= ke λ ln ⎡ x +
⎣
x 2 + b2
V=
ke q ke q ke q
−
=
( r2 − r1 )
r1
r2
r1r2
V = ke
a
P25.61
2 keQ 2
MR
2
a+ L
+ b2 ) ⎤
⎦a
⎡ a + L + ( a + L )2 + b 2
= ke λ ln ⎢
a + a2 + b2
⎢⎣
From the figure, for r >> a,
r2 − r1 ≈ 2a cos θ
Then
v≈
Er = −
k p cos θ
ke q
2a cos θ ≈ e 2
r
r1r2
∂V
2 ke p cos θ
=
∂r
r3
1⎛ ∂ ⎞
In spherical coordinates, the q component of the gradient is − ⎜ ⎟ .
r ⎝ ∂θ ⎠
1 ⎛ ∂V ⎞
ke p sin θ
Therefore,
Eθ = − ⎜ ⎟ =
r ⎝ ∂θ ⎠
r3
For r >> a
continued on next page
⎤
⎥
⎥⎦
Er ( 0°) =
2 ke p
r3
FIG. P25.61
Electric Potential
69
Er ( 90°) = 0,
and
Eθ ( 0° ) = 0
Eθ ( 90° ) =
and
ke p
r3
These results are reasonable for r >> a . Their directions are as shown in Figure 25.13 (c).
However, for r → 0, E ( 0 ) → ∞ . This is unreasonable, since r is not much greater
than a if it is 0.
(c)
V=
(x
ke py
2
and
+ y2 )
32
Ex = −
3ke pxy
∂V
=
∂x ( x 2 + y 2 )5 2
Ey = −
P25.62
2
2
∂V ke p ( 2 y − x )
=
∂y
( x 2 + y2 )5 2
ke q
.
r
The element of charge in a shell is dq = ρ (volume element) or dq = ρ ( 4π r 2 dr ) and the charge
q in a sphere of radius r is
dU = Vdq where the potential V =
r
⎛ 4π r 3 ⎞
q = 4πρ ∫ r 2 dr = ρ ⎜
⎝ 3 ⎟⎠
0
Substituting this into the expression for dU, we have
⎛ 4π r 3 ⎞ ⎛ 1 ⎞
⎛ 16π 2 ⎞ 2 4
⎛ k q⎞
ρ r dr
dU = ⎜ e ⎟ dq = ke ρ ⎜
ρ ( 4π r 2 dr = ke ⎜
⎟
⎜
⎟
⎝ r ⎠
⎝ 3 ⎠ ⎝ r⎠
⎝ 3 ⎟⎠
)
⎛ 16π 2 ⎞ 2 4
⎛ 16π 2 ⎞ 2 5
ρ ∫ r dr = ke ⎜
ρ R
U = ∫ dU = ke ⎜
⎟
⎝ 3 ⎠ 0
⎝ 15 ⎟⎠
R
But the total charge, Q = ρ
P25.63
2
4
π R 3 . Therefore, U = 3 keQ .
3
5 R
For an element of area which is a ring of radius r and width dr, dV =
dq = σ dA = Cr ( 2π rdr ) and
R
V = C ( 2π ke ) ∫
0
ke dq
r2 + x2
⎡
⎛
⎞⎤
x
= C (π ke ) ⎢ R R 2 + x 2 + x 2 ln ⎜
⎥
2
2 ⎟
⎝ R+ R + x ⎠⎦
r +x
⎣
r 2 dr
2
2
.
70
Chapter 25
*P25.64 (a)
keQ/r = 8.99 × 109 (1.6 × 10−9) V/2 = 7.19 V
(b)
8.99 1.6 8.99 1.6
+
= 7.67 V
2(1 + 3 2)
2(1+ 1 2)
(c)
8.99 1.6 ⎛ 1
1
1
1 ⎞
+
+
⎜ 1 +
⎟ = 7.84 V
3
5
4
⎝1 + 4 1 + 4 1 + 4 1 + 7 4 ⎠
(d)
We find
(e)
We find 8.99 1.6 ⎛⎜ 1 + 1 + ... + 1 ⎞⎟ = 7.901 0 V
64 ⎝ 1 + 164 1 + 3 64
1 + 127 64 ⎠
(f)
We represent the exact result as
V=
8.99 1.6 ⎛ 1
1
1 ⎞
= 7.900 2 V
+
+ ... +
32 ⎜⎝ 1 + 1 32 1 + 3 32
1 + 63 32 ⎟⎠
keQ ⎛ + a ⎞ 8.99 1.6 V ⎛ 3 ⎞
ln ⎜
ln ⎜ ⎟ = 7.901 2 V
⎟=
⎝1⎠
⎝ a ⎠
2
Modeling the line as a set of points works nicely. The exact result, represented as 7.901 2 V, is
approximated to within 0.8% by the four-particle version. The 16-particle approximation gives a
result accurate to three digits, to within 0.05%. The 64-charge approximation gives a result accurate to four digits, differing by only 0.003% from the exact result.
P25.65
The positive plate by itself creates a field E =
σ
36.0 × 10 −9 C m 2
=
= 2.03 kN C
2 ∈0 2 ( 8.85 × 10 −12 C2 N ⋅ m 2 )
away from the + plate. The negative plate by itself creates the same size field and between the
plates it is in the same direction. Together the plates create a uniform field 4.07 kN C in the
space between.
(a)
Take V = 0 at the negative plate. The potential at the positive plate is then
12.0 cm
∫ ( −4.07
V −0=−
kN C ) dx
0
The potential difference between the plates is V = ( 4.07 × 10 3 N C) ( 0.120 m ) = 488 V .
(b)
⎛1
⎞
⎞ ⎛1
2
2
⎜⎝ m v + qV ⎟⎠ = ⎜⎝ m v + qV ⎟⎠
2
2
i
f
)
qV = (1.60 × 10 −19 C ( 488 V) =
(c)
v f = 306 km s
(d)
v 2f = vi2 + 2 a ( x f − xi )
( 3.06 × 10
5
ms
)
2
= 0 + 2 a ( 0.120 m )
a = 3.90 × 1011 m s 2
(e)
1
m v 2f = 7.81 × 10 −17 J
2
∑ F = ma = (1.67 × 10
toward the negative plate
−27
)
plate
(f)
E=
)
kg ( 3.90 × 1011 m s 2 = 6.51 × 10 −16 N
F 6.51 × 10 −16 N
=
= 4.07 kN C
q 1.60 × 10 −19 C
toward the negative
Electric Potential
P25.66
71
ke ( q ) ke ( −2q )
+
r1
r2
For the given charge distribution,
V ( x , y, z ) =
where
2
r1 = ( x + R ) + y 2 + z 2 and r2 = x 2 + y 2 + z 2
The surface on which
V ( x , y, z ) = 0
is given by
⎛1 2 ⎞
ke q ⎜ − ⎟ = 0, or 2r1 = r2
⎝ r1 r2 ⎠
This gives:
4 ( x + R) + 4 y2 + 4 z 2 = x 2 + y2 + z 2
which may be written in the form:
8
4
x 2 + y 2 + z 2 + ⎛ R⎞ x + ( 0 ) y + ( 0 ) z + ⎛ R 2 ⎞ = 0
⎝3 ⎠
⎝3 ⎠
2
[1]
The general equation for a sphere of radius a centered at ( x0 , y0 , z0 ) is:
( x − x0 )2 + ( y − y0 )2 + ( z − z0 )2 − a 2 = 0
or
x 2 + y 2 + z 2 + ( −2 x0 ) x + ( −2 y0 ) y + ( −2 z0 ) z + ( x02 + y02 + z02 − a 2 ) = 0
[2]
Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 is
indeed a sphere and that:
4
8
R; − 2 y0 = 0; − 2 z0 = 0; x02 + y02 + z02 − a 2 = R 2
3
3
16
4
4
4
Thus, x0 = − R, y0 = z0 = 0, and a 2 = ⎛ − ⎞ R 2 = R 2
⎝ 9 3⎠
9
3
−2 x0 =
⎛ 4
The equipotential surface is therefore a sphere centered at ⎜− R, 0,
⎝ 3
P25.67
⎞
2
0 ⎟ , having a radius
R .
⎠
3
Inside the sphere, Ex = Ey = Ez = 0 .
(
−3 2
∂V
∂
=−
V0 − E0 z + E0 a 3 z ( x 2 + y 2 + z 2 )
∂x
∂x
)
Outside,
Ex = −
So
−5 2
−5 2
3
⎡
⎤
Ex = − ⎢ 0 + 0 + E0 a 3 z ⎛ − ⎞ ( x 2 + y 2 + z 2 ) ( 2 x ) ⎥ = 3E0 a 3 xz ( x 2 + y 2 + z 2 )
⎝
⎠
2
⎣
⎦
Ey = −
(
−3 2
∂V
∂
=−
V0 − E0 z + E0 a 3 z ( x 2 + y 2 + z 2 )
∂y
∂y
)
−5 2
−5 2
3
Ey = − E0 a 3 z ⎛ − ⎞ ( x 2 + y 2 + z 2 ) 2 y = 3E0 a 3 yz ( x 2 + y 2 + z 2 )
⎝ 2⎠
Ez = −
−3 2
−55 2
∂V
3
= E0 − E0 a 3 z ⎛ − ⎞ ( x 2 + y 2 + z 2 ) ( 2 z ) − E0 a 3 ( x 2 + y 2 + z 2 )
⎝ 2⎠
∂z
E z = E0 + E0 a 3 ( 2 z 2 − x 2 − y 2 ) ( x 2 + y 2 + z 2 )
−5 2
72
Chapter 25
ANSWERS TO EVEN PROBLEMS
P25.2
1.35 MJ
P25.4
+260 V
P25.6
See the solution.
P25.8
0.300 m s
P25.10
(a) 0
P25.12
(a) −4.83 m
(b) 0.667 m and −2.00 m
P25.14
(a) −386 nJ
(b) 103 V
P25.16
(a) 32.2 kV
(b) −96.5 mJ
P25.18
(a) +7.00 nC with position vector 7.00 cm at 70.0° and −8.00 nC with position vector 3.00 cm at
90.0° (b) (0.070 3 î − 1.09 ĵ)mN (c) +24.0 mJ
P25.20
−
P25.22
See the solution.
P25.24
(a) v1 =
(b) 0
(c) 45.0 kV
5ke q
R
2 m2 ke q1q2 ⎛ 1
1⎞
− ⎟
⎜
m1 ( m1 + m2 ) ⎝ r1 + r2 d ⎠
v2 =
2 m1 ke q1q2 ⎛ 1
1⎞
− ⎟
⎜
m2 ( m1 + m2 ) ⎝ r1 + r2 d ⎠
(b) Faster than
calculated in (a)
P25.26
27.4 fm
P25.28
22.8
ke q 2
s
P25.30
(a) 0
(b)
P25.32
(a) larger at A
P25.34
−0.553
P25.36
−
P25.38
kel(p + 2 ln3)
P25.40
1.56 × 1012 electrons
P25.42
(a) 135 kV
sphere
P25.44
See the solution.
keQ
radially outward
r2
(b) 200 N C down
(c) See the solution.
keQ
R
2
2
keα L ⎡⎢ b + ( L 4 ) − L 2 ⎤⎥
ln
2
⎢ b 2 + ( L2 4 ) + L 2 ⎥
⎣
⎦
(b) 2.25 MV m away from the large sphere and 6.74 MV m away from the small
Electric Potential
73
P25.46
(a) The velocity of one particle relative to the other is first a velocity of approach, then zero at
closest approach, and then a velocity of recession. (b) 6.00 î m/s (c) 3.64 m (d) −9.00 î m/s
for the incident particle and 12.0 î m/s for the target particle.
P25.48
(a) negative 1.06 nC/m2 (b) −542 kC (c) 764 MV (d) His head is higher in potential by
210 V. (e) 4.88 kN away from the Earth (f) The gravitational force is in the opposite direction
and 4.08 × 1016 times larger. Electrical forces are negligible in accounting for planetary motion.
(g) −490 nC (h) Less charge to be suspended at the equator. The gravitational force is weaker
at a greater distance from the Earth’s center. The suspended particle is not quite in equilibrium,
but accelerating downward to participate in the daily rotation. At uniform potential, the
planet’s surface creates a stronger electric field at the equator, where its radius of curvature is
smaller.
P25.50
(a) ~10 4 V
P25.52
See the solution.
P25.54
(b) ~10 −6 C
2
2 ⎞
⎛
(a) keQ ln ⎜ d + h + ( d + h ) + R ⎟
h
d + d 2 + R2
⎠
⎝
(b)
⎛ d + h + ( d + h )2 + R 2 ⎞ ⎤
keQ ⎡
⎢( d + h ) ( d + h )2 + R 2 − d d 2 + R 2 − 2dh − h 2 + R 2 ln ⎜
2
⎟⎥
Rh ⎢
d + d 2 + R2
⎠ ⎥⎦
⎝
⎣
P25.56
(a) −keq/4 a (b) The approximate expression −2keqa/x2 gives −keq/4.5 a, which is different by
only 11.1%.
P25.58
⎛ ke q 2 ⎞
⎜⎝ 3am ⎟⎠
P25.60
⎡ a + L + ( a + L )2 + b 2 ⎤
⎥
ke λ ln ⎢
a + a2 + b2
⎢⎣
⎥⎦
P25.62
3 keQ 2
5 R
12
P25.64
(a) 7.19 V (b) 7.67 V (c) 7.84 V (d) 7.900 2 V (e) 7.901 0 V (f) Modeling the line as a
set of points works nicely. The exact result, represented as 7.901 2 V, is approximated to within
0.8% by the four-particle version. The 16-particle approximation gives a result accurate to three
digits, to within 0.05%. The 64-charge approximation gives a result accurate to four digits,
differing by only 0.003% from the exact result.
P25.66
See the solution.
26
Capacitance and Dielectrics
CHAPTER OUTLINE
26.1
26.2
26.3
26.4
26.5
26.6
26.7
ANSWERS TO QUESTIONS
Definition of Capacitance
Calculating Capacitance
Combinations of Capacitors
Energy Stored in a Charged
Capacitor
Capacitors with Dielectrics
Electric Dipole in an Electric Field
An Atomic Description of
Dielectrics
*Q26.1 (a) False. (b) True. In Q = C∆V, the capacitance is the
proportionality constant relating the variables Q and ∆V.
Q26.2
Seventeen combinations:
C1 , C2 , C3
Individual
C1 + C2 + C3 , C1 + C2 , C1 + C3 , C2 + C3
Parallel
−1
−1
−1
Series-Parallel
⎛ 1
⎛ 1
⎛ 1
1⎞
1⎞
1⎞
⎜⎝ C + C ⎟⎠ + C3 , ⎜⎝ C + C ⎟⎠ + C2 , ⎜⎝ C + C ⎟⎠ + C1
1
3
2
3
1
2
−1
−1
⎛ 1
⎛ 1
⎛ 1
1⎞
1⎞
1⎞
⎜⎝ C + C + C ⎟⎠ , ⎜⎝ C + C + C ⎟⎠ , ⎜⎝ C + C + C ⎟⎠
2
3
1
1
2
3
1
3
2
−1
−1
Series
−1
−1
⎛ 1
⎛ 1
⎛ 1
⎛ 1
1⎞
1⎞
1⎞
1
1⎞
⎜⎝ C + C + C ⎟⎠ , ⎜⎝ C + C ⎟⎠ , ⎜⎝ C + C ⎟⎠ , ⎜⎝ C + C ⎟⎠
2
3
1
3
1
2
3
1
2
−1
*Q26.3 Volume is proportional to radius cubed. Increasing the radius by a factor of 31/3 will triple the
volume. Capacitance is proportional to radius, so it increases by a factor of 31/3 . Answer (d).
*Q26.4 Let C2 = NC1 be the capacitance of the large capacitor and C1 that of the small one.
The equivalent capacitance is
Ceq = Ceq =
1
1
N
=
=
C1
1 / C1 + 1 / NC1 ( N + 1) / NC1 N + 1
This is slightly less than C1, answer (d).
*Q26.5 We find the capacitance, voltage, charge, and energy for each capacitor.
∆V = 4 V
Q = C∆V = 80 µC U = (1/2)Q∆V = 160 µJ
(a) C = 20 µF
(b) C = 30 µF
∆V = Q/C = 3 V
Q = 90 µC
U = 135 µJ
(c) C = Q/∆V = 40 µF ∆V = 2 V
Q = 80 µC
U = 80 µJ
(d) C = 10 µF
∆V = (2U/C)1/2 = 5 V Q = 50 µC
U = 125 µJ
(e) C = 2U/∆V 2 = 5 µF ∆V = 10 V
Q = 50 µC
U = 250 µJ
(f) C = Q2/2U = 20 µF ∆V = 6 V
Q = 120 µC
U = 360 µJ
Then (i) the ranking by capacitance is c > b > a = f > d > e.
(ii) The ranking by voltage ∆V is e > f > d > a > b > c.
(iii) The ranking by charge Q is f > b > a = c > d = e.
(iv) The ranking by energy U is f > e > a > b > d > c.
75
76
Q26.6
Chapter 26
A capacitor stores energy in the electric field between the plates. This is most easily seen when
using a “dissectible” capacitor. If the capacitor is charged, carefully pull it apart into its component pieces. One will find that very little residual charge remains on each plate. When reassembled, the capacitor is suddenly “recharged”—by induction—due to the electric field set up and
“stored” in the dielectric. This proves to be an instructive classroom demonstration, especially
when you ask a student to reconstruct the capacitor without supplying him/her with any rubber
gloves or other insulating material. (Of course, this is after they sign a liability waiver.)
*Q26.7 (i) According to Q = C∆V, the answer is (b).
(ii) From U = (1/2)C∆V 2, the answer is (a).
*Q26.8 The charge stays constant as C is cut in half, so U = Q2/2C doubles: answer (b).
*Q26.9 (i) Answer (b).
(ii) Answer (c).
(iii) Answer (c).
(iv) Answer (a).
(v) Answer (a).
*Q26.10 (i) Answer (b).
(ii) Answer (b).
(iii) Answer (b).
(iv) Answer (c).
(v) Answer (b).
Q26.11
The work you do to pull the plates apart becomes additional electric potential energy stored in
the capacitor. The charge is constant and the capacitance decreases but the potential difference
1
increases to drive up the potential energy Q ∆V . The electric field between the plates is
2
constant in strength but fills more volume as you pull the plates apart.
Q26.12 The work done, W = Q ∆V , is the work done by an external agent, like a battery, to move a charge
through a potential difference, ∆V. To determine the energy in a charged capacitor, we must add
the work done to move bits of charge from one plate to the other. Initially, there is no potential
difference between the plates of an uncharged capacitor. As more charge is transferred from one
plate to the other, the potential difference increases as shown in the textbook graph
of ∆V versus Q, meaning that more work is needed to transfer each additional bit of charge.
1
The total work is the area under the curve on this graph, and thus W = Q ∆V .
2
*Q26.13 Let C = the capacitance of an individual capacitor, and Cs represent the equivalent capacitance
of the group in series. While being charged in parallel, each capacitor receives charge
Q = C ∆Vcharge = ( 5.00 × 10 −4 F ) (800 V ) = 0.400 C
While being discharged in series,
∆Vdischarge =
0.400 C
Q
Q
=
=
= 8.00 kV
Cs C 10 5.00 × 10 −5 F
(or 10 times the original voltage). Answer (b).
Q26.14 The potential difference must decrease. Since there is no external power supply, the charge on
the capacitor, Q, will remain constant—that is, assuming that the resistance of the meter is
sufficiently large. Adding a dielectric increases the capacitance, which must therefore decrease
the potential difference between the plates.
*Q26.15 (i) Answer (a).
(iii) Answer (c).
(ii) Because ∆V is constant, Q = C∆V increases, answer (a).
(iv) Answer (c).
(v) U = (1/2)C∆V 2 increases, answer (a).
Q26.16 Put a material with higher dielectric strength between the plates, or evacuate the space between
the plates. At very high voltages, you may want to cool off the plates or choose to make them of a
different chemically stable material, because atoms in the plates themselves can ionize, showing
thermionic emission under high electric fields.
Capacitance and Dielectrics
77
Q26.17 The primary choice would be the dielectric. You would want to choose a dielectric that has a large
dielectric constant and dielectric strength, such as strontium titanate, where κ ≈ 233 (Table 26.1).
A convenient choice could be thick plastic or Mylar. Secondly, geometry would be a factor. To maximize capacitance, one would want the individual plates as close as possible, since the capacitance is
proportional to the inverse of the plate separation—hence the need for a dielectric with a high dielectric
strength. Also, one would want to build, instead of a single parallel plate capacitor, several capacitors
in parallel. This could be achieved through “stacking” the plates of the capacitor. For example, you can
alternately lay down sheets of a conducting material, such as aluminum foil, sandwiched between your
sheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their
next neighbors, connect every other plate together. Figure Q26.17 illustrates this idea.
Dielectric
Conductor
FIG. Q26.17
This technique is often used when “home-brewing” signal capacitors for radio applications, as
they can withstand huge potential differences without flashover (without either discharge between
plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich
together flexible materials such as aluminum roof flashing and thick plastic, so the whole product
can be rolled up into a “capacitor burrito” and placed in an insulating tube, such as a PVC pipe, and
then filled with motor oil (again to prevent flashover).
SOLUTIONS TO PROBLEMS
Section 26.1
P26.1
P26.2
)
F ) (1.50 V) = 6.00 × 10
(a)
Q = C ∆V = ( 4.00 × 10 −6 F (12.0 V) = 4.80 × 10 −5 C = 48.0 µC
(b)
Q = C ∆V = ( 4.00 × 10 −6
(a)
C=
(b)
∆V =
Section 26.2
P26.3
Definition of Capacitance
−6
C = 6.00 µC
Q 10.0 × 10 −6 C
=
= 1.00 × 10 −6 F = 1.00 µ F
∆V
10.0 V
Q 100 × 10 −6 C
=
= 100 V
C 1.00 × 10 −6 F
Calculating Capacitance
kq
E = e2 :
r
( 4.90 × 10 N C) ( 0.210 m )
q=
( 8.99 × 10 N ⋅ m C )
4
2
9
2
2
= 0.240 µC
q 0.240 × 10 −6
=
= 1.33 µC m 2
A 4π ( 0.120 )2
(a)
σ=
(b)
C = 4π ∈0 r = 4π ( 8.85 × 10 −12 ( 0.120 ) = 13.3 pF
)
78
Chapter 26
3
−12
2
κ ∈0 A (1.00 ) ( 8.85 × 10 C ) (1.00 × 10 m )
C=
=
= 11.1 nF
d
N ⋅ m 2 ( 800 m )
2
P26.4
The potential between ground and cloud is
)
∆V = Ed = ( 3.00 × 10 6 N C ( 800 m ) = 2.40 × 10 9 V
Q = C ( ∆V ) = (11.1 × 10
P26.5
P26.6
)
20.0 V
= 11.1 kV m
1.80 × 10 −3 m
(a)
∆V = Ed so
(b)
E=
(c)
∈ A ( 8.85 × 10
C= 0 =
d
(d)
∆V =
σ
so
∈0
E=
)
C V ( 2.40 × 10 9 V = 26.6 C
−9
toward the negative plate
σ = (1.11 × 10 4 N C ) ( 8.85 × 10 −12 C2 N ⋅ m 2 ) = 98.3 nC m 2
Q
so
C
−12
)
)
C2 N ⋅ m 2 ( 7.60 cm 2 (1.00 m 100 cm )
1.80 × 10 −3 m
2
= 3.74 pF
)
Q = ( 20.0 V) ( 3.74 × 10 −12 F = 74.7 pC
With θ = π , the plates are out of mesh and the overlap area is zero.
π R2
. By proportion,
2
(π − θ ) R 2
.
the effective area of a single sheet of charge is
2
With θ = 0, the overlap area is that of a semi-circle,
When there are two plates in each comb, the number of adjoining
sheets of positive and negative charge is 3, as shown in the sketch. When
there are N plates on each comb, the number of parallel capacitors
is 2 N − 1 and the total capacitance is
C = ( 2 N − 1)
P26.7
P26.8
∈0 Aeffective ( 2 N − 1) ∈0 (π − θ ) R 2 2
( 2 N − 1) ∈0 (π − θ ) R 2
=
.
=
distance
d 2
d
∈ ( ∆V )
Q
=σ = 0
A
d
Q=
∈0 A
( ∆V )
d
d=
( 8.85 × 10−12 C2 N ⋅ m 2 (150 V) = 4.42 µm
∈0 ( ∆V )
=
σ
( 30.0 × 10−9 C cm 2 (1.00 × 10 4 cm 2 m 2
)
∑F
∑F
y
= 0:
T cos θ − mg = 0
x
= 0:
T sin θ − Eq = 0
Eq
mg
Dividing,
tan θ =
so
E=
and
∆V = Ed =
mg
tan θ
q
mgd tan θ
q
)
)
FIG. P26.6
Capacitance and Dielectrics
P26.9
50.0
=
= 2.68 nF
9
2 ke ln ( b / a ) 2 ( 8.99 × 10 ln ( 7.27 / 2.58 )
(a)
C=
(b)
Method 1:
)
⎛ b⎞
∆V = 2 ke λ ln ⎜ ⎟
⎝ a⎠
q 8.10 × 10 −6 C
=
= 1.62 × 10 −7 C m
50.0 m
⎛ 7.27 ⎞
= 3.02 kV
∆V = 2 ( 8.99 × 10 9 (1.62 × 10 −7 ln ⎜
⎝ 2.58 ⎟⎠
λ=
)
Method 2:
∆V =
)
Q 8.10 × 10 −6
=
= 3.02 kV
C 2.68 × 10 −9
*P26.10 The original kinetic energy of the particle is
2
1
1
K = m v 2 = ( 2 × 10 −16 kg ) ( 2 × 10 6 m s ) = 4.00 × 10 −4 J
2
2
Q 1 000 µC
The potential difference across the capacitor is ∆V = =
= 100 V.
C
10 µ F
For the particle to reach the negative plate, the particle-capacitor system would need energy
U = q∆V = ( −3 × 10 −6 C ) ( −100 V ) = 3.00 × 10 −4 J
Since its original kinetic energy is greater than this, the particle will reach the negative plate .
As the particle moves, the system keeps constant total energy
4.00 × 10 −4 J + (−3 × 10 −6 C) (+100 V) =
( K + U )at + plate = ( K + U )at − plate:
2 (1.00 × 10 −4 J )
vf =
P26.11
= 1.00 × 10 6 m s
(a)
C=
( 0.070 0 ) ( 0.140 )
ab
=
= 15.6 pF
ke ( b − a ) ( 8.99 × 10 9 ( 0.140 − 0.070 0 )
(b)
C=
Q
∆V
Section 26.3
P26.12
2 × 10 −16 kg
(a)
1
(2 × 10−16 ) v2f + 0
2
)
∆V =
Q 4.00 × 10 −6 C
=
= 256 kV
C 15.6 × 10 −12 F
Combinations of Capacitors
Capacitors in parallel add. Thus, the equivalent capacitor has a value of
Ceq = C1 + C2 = 5.00 µ F + 12.0 µ F = 17.0 µ F
(b)
The potential difference across each branch is the same and equal to the voltage of the battery.
∆V = 9.00 V
(c)
Q5 = C ∆V = ( 5.00 µ F ) ( 9.00 V) = 45.0 µC
and Q12 = C ∆V = (12.0 µ F ) ( 9.00 V) = 108 µC
79
80
P26.13
Chapter 26
(a)
(c)
In series capacitors add as
1
1
1
1
1
=
+
=
+
Ceq C1 C2 5.00 µ F 12.0 µ F
and
Ceq = 3.53 µ F
The charge on the equivalent capacitor is
Qeq = Ceq ∆V = ( 3.53 µ F ) ( 9.00 V ) = 31.8 µ C
Each of the series capacitors has this same charge on it.
(b)
P26.14
So
Q1 = Q2 = 31.8 µ C
The potential difference across each is
∆V1 =
Q1 31.8 µC
=
= 6.35 V
C1 5.00 µ F
and
∆V2 =
Q2 31.8 µ C
=
= 2.65 V
C2 12.0 µ F
(a)
Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in
−1
1 ⎤
⎡ 1
series with capacitor 1, so the battery sees capacitance ⎢
+
= 2C .
⎣ 3C 6C ⎥⎦
(b)
If they were initially uncharged, C1 stores the same charge as C2 and C3 together.
With greater capacitance, C3 stores more charge than C2 . Then Q1 > Q3 > Q2 .
(c)
(d)
The ( C2 || C3 ) equivalent capacitor stores the same charge as C1. Since it has greater
Q
capacitance, ∆V = implies that it has smaller potential difference across it than C1.
C
In parallel with each other, C2 and C3 have equal voltages: ∆V1 > ∆V2 = ∆V3 .
If C3 is increased, the overall equivalent capacitance increases. More charge moves
through the battery and Q increases. As ∆V1 increases, ∆V2 must decrease so Q2 decreases.
Then Q3 must increase even more: Q3 and Q1 increase; Q2 decreases .
P26.15
C p = C1 + C2
1
1
1
=
+
Cs C1 C2
Substitute C2 = C p − C1
C p − C1 + C1
1
1
1
=
+
=
Cs C1 C p − C1 C1 C p − C1
Simplifying,
C12 − C1C p + C p Cs = 0
C1 =
C p ± C p2 − 4 C p Cs
2
(
=
)
1
1 2
Cp ±
C p − C p Cs
2
4
We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus
sign, we would get the same two answers with their names interchanged.)
C1 =
1
1 2
1
1
Cp +
C p − C p Cs = ( 9.00 pF ) +
( 9.00 pF )2 − ( 9.00 pF ) ( 2.00 pF ) = 6.00 pF
2
4
2
4
C2 = C p − C1 =
1
1 2
1
C p − C p Cs = ( 9.00 pF ) − 1.50 pF = 3.00 pF
Cp −
2
4
2
Capacitance and Dielectrics
P26.16
81
C p = C1 + C2
and
1
1
1
=
+ .
Cs C1 C2
Substitute
C2 = C p − C1 :
C p − C1 + C1
1
1
1
=
+
=
Cs C1 C p − C1 C1 C p − C1
)
(
Simplifying,
C12 − C1C p + C p Cs = 0
and
C1 =
C p ± C p2 − 4 C p Cs
2
=
1
1 2
Cp +
C p − C p Cs
2
4
where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values
for the capacitances, with the names reversed).
Then, from
C2 = C p − C1
C2 =
P26.17
(a)
1
1 2
Cp −
C p − C p Cs
2
4
1
1
1
=
+
Cs 15.0 3.00
Cs = 2.50 µ F
C p = 2.50 + 6.00 = 8.50 µ F
⎛
1
1 ⎞
Ceq = ⎜
+
⎝ 8.50 µ F 20.0 µ F ⎟⎠
(b)
−1
= 5.96 µ F
Q = C ∆V = ( 5.96 µ F ) (15.0 V) = 89.5 µC on 20.0 µ F
Q 89.5 µC
=
= 4.47 V
C 20.0 µ F
15.0 − 4.47 = 10.53 V
∆V =
Q = C ∆V = ( 6.00 µ F ) (10.53 V) = 63.2 µC on 6.00 µ F
89.5 − 63.2 = 26.3 µC on 15.0 µ F and 3.00 µ F
P26.18
(a)
In series , to reduce the effective capacitance:
1
1
1
=
+
32.0 µ F 34.8 µ F Cs
1
Cs =
= 398 µ F
2.51 × 10 −3 µ F
(b)
In parallel , to increase the total capacitance:
29.8 µ F + C p = 32.0 µ F
C p = 2.20 µ F
FIG. P26.17
82
P26.19
Chapter 26
C=
Q
so
∆V
and
6.00 × 10 −6 =
Q
20.0
Q = 120 µC
Q1 = 120 µC − Q2
and
or
120 − Q2 Q2
=
C1
C2
Q
:
C
120 − Q2
Q
= 2
6.00
3.00
∆V =
FIG. P26.19
( 3.00 ) (120 − Q2 ) = ( 6.00 ) Q2
Q2 =
360
= 40.0 µC
9.00
Q1 = 120 µC − 40.0 µC = 80.0 µC
P26.20
For C1 connected by itself, C1 ∆V = 30.8 µC where ∆V is the battery voltage: ∆V =
For C1 and C2 in series:
⎛
⎞
1
⎜⎝ 1 C + 1 C ⎟⎠ ∆V = 23.1 µC
1
2
substituting, 30.8 µC = 23.1 µC + 23.1 µC
C1
C1
C2
C1 = 0.333C2
For C1 and C3 in series:
⎛
⎞
1
⎜⎝ 1 C + 1 C ⎟⎠ ∆V = 25.2 µC
1
3
30.8 µC 25.2 µC 25.2 µC
=
+
C1
C1
C3
C1 = 0.222C3
For all three:
⎛
⎞
1
C1 ∆V
30.8 µ C
Q=⎜
∆V =
=
= 19.8 µ C
⎟
1 + C1 C2 + C1 C3 1 + 0.333 + 0.222
⎝ 1 C1 + 1 C2 + 1 C3 ⎠
This is the charge on each one of the three.
P26.21
nC =
nC =
100
1
1 + 1 +
C+
C
C n capacitors
100C
n
so
=
100
nC
n 2 = 100
and
n = 10
30.8 µ C
.
C1
Capacitance and Dielectrics
P26.22
According to the suggestion, the combination
of capacitors shown is equivalent to
Then
1
1
1
1
=
+
+
C C0 C + C0 C0
C + C0 + C0 + C + C0
=
C0 (C + C0 )
C0 C + C02 = 2C 2 + 3C0 C
2C 2 + 2C0 C − C02 = 0
C=
−2C0 ± 4 C02 + 4 ( 2C02
)
FIG. P26.22
4
Only the positive root is physical.
C=
P26.23
C0
2
(
1 ⎞
⎛ 1
Cs = ⎜
+
⎝ 5.00 10.0 ⎟⎠
)
3 −1
−1
= 3.33 µ F
C p1 = 2 ( 3.33) + 2.00 = 8.66 µ F
C p 2 = 2 (10.0 ) = 20.0 µ F
1 ⎞
⎛ 1
Ceq = ⎜
+
⎝ 8.66 20.0 ⎟⎠
−1
= 6.04 µ F
FIG. P26.23
P26.24
)
Qeq = Ceq ( ∆V ) = ( 6.04 × 10 −6 F ( 60.0 V) = 3.62 × 10 −4 C
Q p1 = Qeq ,
(
so
∆Vp1 =
)
Qeq
C p1
=
3.62 × 10 −4 C
= 41.8 V
8.66 × 10 −6 F
)
Q3 = C3 ∆Vp1 = ( 2.00 × 10 −6 F ( 41.8 V) = 83.6 µC
P26.25
1 ⎞
⎛ 1
Cs = ⎜
+
⎝ 5.00 7.00 ⎟⎠
−1
= 2.92 µ F
C p = 2.92 + 4.00 + 6.00 = 12.9 µ F
FIG. P26.25
83
84
Chapter 26
Section 26.4
P26.26
U=
Energy Stored in a Charged Capacitor
1
C ∆V 2
2
∆V =
P26.27
P26.28
2 ( 300 J )
= 4.47 × 10 3 V
30 × 10 −6 C V
2U
=
C
(a)
U=
1
1
2
2
C ( ∆V ) = ( 3.00 µ F ) (12.0 V) = 216 µ J
2
2
(b)
U=
1
1
2
2
C ( ∆V ) = ( 3.00 µ F ) ( 6.00 V) = 54.0 µ J
2
2
U=
1
2
C ( ∆V )
2
The circuit diagram is shown at the right.
(a)
C p = C1 + C2 = 25.0 µ F + 5.00 µ F = 30.0 µ F
U=
(b)
)
1
2
30.0 × 10 −6 (100 ) = 0.150 J
(
2
⎛ 1
1⎞
Cs = ⎜ + ⎟
⎝ C1 C2 ⎠
−1
⎛
1
1 ⎞
=⎜
+
⎝ 25.0 µ F 5.00 µ F ⎟⎠
−1
= 4.17 µ F
FIG. P26.28
1
2
U = C ( ∆V )
2
∆V =
P26.29
2 ( 0.150 )
= 268 V
4.17 × 10 −6
W = U = ∫ Fdx
so
P26.30
2U
=
C
F=
dU d ⎛ Q 2 ⎞ d ⎛ Q 2 x ⎞
Q2
= ⎜ ⎟= ⎜
⎟=
dx dx ⎝ 2C ⎠ dx ⎝ 2 ∈ 0 A ⎠ 2 ∈ 0 A
With switch closed, distance d ′ = 0.500 d and capacitance C ′ =
∈0 A 2 ∈0 A
=
= 2C .
d′
d
)
(a)
Q = C ′ ( ∆V ) = 2C ( ∆V ) = 2 ( 2.00 × 10 −6 F (100 V) = 400 µC
(b)
The force stretching out one spring is
Q2
4C 2 ( ∆V )
2C 2 ( ∆V )
2C ( ∆V )
=
=
=
2∈0 A
2∈0 A
d
(∈0 A d ) d
2
F=
2
2
d
One spring stretches by distance x = , so
4
2
2
8 ( 2.00 × 10 −6 F ) (100 V )
F 2C ( ∆V ) ⎛ 4 ⎞ 8C ( ∆V )
=
=
=
= 2.50 kN m
⎝ d⎠
x
d
d2
(8.00 × 10 −3 m )2
2
k=
Capacitance and Dielectrics
P26.31
)
Q = C ∆V = (150 × 10 −12 F (10 × 10 3 V = 1.50 × 10 −6 C
(b)
U=
1
2
C ( ∆V )
2
∆V =
P26.32
)
(a)
2U
=
C
2 ( 250 × 10 −6 J
150 × 10 −12 F
)=
1
1
2
2
2
C ( ∆V ) + C ( ∆V ) = C ( ∆V )
2
2
(a)
U=
(b)
The altered capacitor has capacitance C ′ =
C ( ∆V ) + C ( ∆V ) = C ( ∆V ′ ) +
C
( ∆V ′ )
2
C
. The total charge is the same as before:
2
∆V ′ =
( ∆V )
1 ⎛ 4 ∆V ⎞
1 1 ⎛ 4 ∆V ⎞
= 4C
+
C⎜
C⎜
⎟
⎟
2 ⎝ 3 ⎠
22 ⎝ 3 ⎠
3
2
P26.33
1.83 × 10 3 V
2
4 ∆V
3
2
(c)
U′ =
(d)
The extra energy comes from work put into the system by the agent pulling the capacitor
plates apart.
U=
1
R
kQ
kQ
2
C ( ∆V ) where C = 4π ∈0 R =
and ∆V = e − 0 = e
2
ke
R
R
U=
1 ⎛ R ⎞ ⎛ keQ ⎞
keQ 2
=
⎟
⎜
2 ⎜⎝ ke ⎟⎠ ⎝ R ⎠
2R
2
1 q12 1 q22 1
1 (Q − q1 )
q12
.
+
=
+
2 C1 2 C2 2 4 π ∈ 0 R1 2 4 π ∈ 0 R 2
2
P26.34
(a)
The total energy is U = U1 + U 2 =
For a minimum we set
dU
= 0:
dq1
1 2 q1
1 2(Q − q1 )
+
(−1) = 0
2 4 π ∈ 0 R1 2 4 π ∈ 0 R2
R2 q1 = R1Q − R1q1
Then q2 = Q − q1 =
4
(b)
q1 =
R1Q
R1 + R2
R2Q
= q2 .
R1 + R2
V1 =
ke q1
ke R1Q
kQ
=
= e
R1
R1 ( R1 + R2 ) R1 + R2
V2 =
ke q2
ke R2Q
kQ
=
= e
R2
R2 ( R1 + R2 ) R1 + R2
and V1 − V2 = 0
85
86
P26.35
Chapter 26
)
then
∆Eint = m ( 4 186 J kg ⋅ °C ) (100°C − 30.0°C ) + m ( 2.26 × 10 6 J kg = 2.50 × 10 7 J
giving
m = 9.79 kg
Section 26.5
P26.36
)
1
1
TET = Q ∆V = ( 50.0 C ) (1.00 × 10 8 V = 2.50 × 10 9 J
2
2
and 1% of this (or ∆Eint = 2.50 × 10 7 J) is absorbed by the tree. If m is the amount of water boiled
away,
The energy transferred is
Capacitors with Dielectrics
Qmax = C ∆Vmax ,
but
∆Vmax = Emax d
Also,
C=
κ ∈0 A
d
κ ∈0 A
Qmax =
( Emax d ) = κ ∈0 AEmax
d
Thus,
(a)
With air between the plates, κ = 1.00
Emax = 3.00 × 10 6 V m
and
Therefore,
Qmax = κ ∈0 AEmax = (8.85 × 10 −12 F m ) ( 5.00 × 10 −4 m 2 ) ( 3.00 × 10 6 V m ) = 13.3 nC
(b)
With polystyrene between the plates, κ = 2.56 and Emax = 24.0 × 10 6 V m .
Qmax = κ ∈ 0 AEmax = 2.56 ( 8.85 × 10 −12 F m ) ( 5.00 × 10 −4 m 2 ) ( 24.0 × 10 6 V m ) = 272 nC
P26.37
P26.38
−12
−4
2
κ ∈ 0 A 2.10 ( 8.85 × 10 F m ) (1.75 × 10 m )
= 8.13 × 10 −11 F = 81.3 pF
=
d
4.00 × 10 −5 m
(a)
C=
(b)
∆Vmax = Emax d = ( 60.0 × 10 6 V m ( 4.00 × 10 −5 m = 2.40 kV
)
Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between
2.54 cm
them. Suppose the plastic has κ ≈ 3, Emax ~ 10 7 V m , and thickness 1 mil =
. Then,
1 000
−12
2
2
2
κ ∈ 0 A 3 ( 8.85 × 10 C N ⋅ m )( 0.4 m )
~
C=
~ 10 −6 F
2.54 × 10 −5 m
d
∆Vmax = Emax d ~ (10 7 V m ) ( 2.54 × 10 −5 m ) ~ 10 2 V
P26.39
)
C=
or
κ ∈0 A
d
−9
95.0 × 10 =
= 1.04 m
)
3.70 ( 8.85 × 10 −12 ( 0.070 0 ) 0.025 0 × 10 −3
Capacitance and Dielectrics
P26.40
C=
Originally,
(a)
The charge is the same before and after immersion, with value Q =
Q=
(b)
∈0 A
Q
=
d
( ∆V )i
Finally,
Q
κ ∈0 A
Cf =
=
d
( ∆V ) f
( 8.85 × 10
Cf =
−12
)
∈0 A ( ∆V )i
.
d
)
C2 N ⋅ m 2 ( 25.0 × 10 −4 m 2 ( 250 V)
(1.50 × 10−2 m
)
)
80.0 ( 8.85 × 10 −12 C2 N ⋅ m 2 ( 25.0 × 10 −4 m 2
( ∆V ) f
(1.50 × 10
=
−2
m
)
)=
= 369 pC
118 pF
∈ A ( ∆V )i d ( ∆V )i 250 V
Qd
= 0
=
=
= 3.12 V
κ ∈0 A
κ ∈0 Ad
κ
80.00
Originally,
∈ A ( ∆V )i
1
2
U i = C ( ∆V )i = 0
2
2d
Finally,
κ ∈0 A ( ∆V )i ∈0 A ( ∆V )i
1
2
U f = C f ( ∆V ) f =
=
2
2dκ 2
2dκ
So,
∆U = U f − U i =
2
(c)
2
2
− ∈0 A ( ∆V )i (κ − 1)
2dκ
2
∆U = −
P26.41
(8.85 × 10
−12
C2 N ⋅ m 2 ) ( 25.0 × 10 −4 m 2 ) ( 250 V ) ( 79.0 )
2
2 (1.50 × 10 −2 m ) (80.0 )
= − 45.5 nJ
The given combination of capacitors is equivalent to the circuit
diagram shown to the right.
Put charge Q on point A. Then,
Q = ( 40.0 µ F ) ∆VAB = (10.0 µ F ) ∆VBC = ( 40.0 µ F ) ∆VCD
FIG. P26.41
So, ∆VBC = 4 ∆VAB = 4 ∆VCD, and the center capacitor will break down first, at ∆VBC = 15.0 V.
When this occurs,
1
∆VAB = ∆VCD = ( ∆VBC ) = 3.75 V
4
and VAD = VAB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V .
87
88
Chapter 26
Section 26.6
P26.42
(a)
Electric Dipole in an Electric Field
The displacement from negative to positive charge is
)
(
)
(
)
(
2a = −1.20 ˆi + 1.10 ˆj mm − 1.40 ˆi − 1.30 ˆj mm = −2.60 ˆi + 2.40 ˆj × 10 −3 m
The electric dipole moment is
p = 2aq = ( 3.50 × 10 −9 C ) −2.60 ˆi + 2.40 ˆj × 10 −3 m =
)
(
(b)
−12
C⋅m
τ = p × E = ⎡⎣ −9.10 ˆi + 8.40 ˆj × 10 −12 C ⋅ m ⎤⎦ × ⎡⎣ 7.80 ˆi − 4.90 ˆj × 10 3 N C ⎤⎦
τ = +44.6 kˆ − 65.5 kˆ × 10 −9 N ⋅ m = −2.09 × 10 −8 N ⋅ mkˆ
)
(
(
(c)
( −9.10 ˆi + 8.40 ˆj) × 10
)
(
)
U = − p ⋅ E = − ⎡ −9.10 ˆi + 8.40 ˆj × 10 −12 C ⋅ m ⎤ ⋅ ⎡ 7.80 ˆi − 4.90 ˆj × 10 3 N C ⎤
⎣
⎦ ⎣
⎦
)
(
)
(
U = ( 71.0 + 41.2 ) × 10 −9 J = 112 nJ
(d)
2
2
p = ( 9.10 ) + (8.40 ) × 10 −12 C ⋅ m = 12.4 × 10 −12 C ⋅ m
2
2
E = ( 7.80 ) + ( 4.90 ) × 10 3 N C = 9.21 × 10 3 N C
U min = −114 nJ
U max = p E = 114 nJ,
U max − U min = 228 nJ
P26.43
(a)
Let x represent the coordinate of the negative
charge. Then x + 2 a cos θ is the coordinate of the
positive charge. The force on the negative charge
is F− = −qE ( x ) ˆi. The force on the positive charge is
dE
F+ = + qE ( x + 2a cos θ ) ˆi ≈ qE ( x ) ˆi + q
( 2a cos θ ) ˆi .
dx
F+
p
F-
θ
E
FIG. P26.43(a)
dE
dE
The force on the dipole is altogether F = F− + F+ = q
( 2a cos θ ) ˆi = p cos θ ˆi .
dx
dx
(b)
The balloon creates field along the x-axis of
Thus,
dE (−2 ) ke q
=
.
dx
x3
At x = 16.0 cm,
ke q ˆ
i.
x2
9
−6
dE (−2 ) ( 8.99 × 10 ) ( 2.00 × 10 )
=
= −8.78 MN C ⋅ m
3
dx
(0.160)
F = ( 6.30 × 10 −9 C ⋅ m ) (−8.78 × 10 6 N C ⋅ m ) cos 0°ˆi = −55.3iˆ mN
Capacitance and Dielectrics
Section 26.7
P26.44
(a)
An Atomic Description of Dielectrics
Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A ′ << A
parallel to the sheet. The side wall of the cylinder passes no flux of electric field since this
surface is everywhere parallel to the field. Gauss’s law becomes
EA′ + EA′ =
(b)
Q
Q
A′, so E =
directed away from the positive sheet.
∈A
2∈ A
Q
away from the positive and
2∈ A
toward the negative sheet. Together, they create a field of
In the space between the sheets, each creates field
Q
∈A
E=
(c)
Assume that the field is in the positive x-direction. Then, the potential of the positive plate
relative to the negative plate is
+ plate
Qd
Q ˆ ˆ
E
s
i ⋅ − i dx = +
d
⋅
=
−
∫
∫
∈A
∈A
− plate
− plate
+ plate
∆V = −
(d)
P26.45
89
Capacitance is defined by: C =
(
)
Q
Q
∈ A κ ∈0 A
.
=
=
=
d
d
∆V Qd ∈ A
Emax occurs at the inner conductor’s surface.
2 ke λ
from an equation derived about this situation in Chapter 24.
a
b
∆V = 2 ke λ ln ⎛ ⎞ from Example 26.1.
⎝ a⎠
∆V
Emax =
a ln ( b a )
Emax =
3.00 ⎞
b
= 19.0 kV
∆Vmax = Emax a ln ⎛ ⎞ = (18.0 × 10 6 V m ) ( 0.800 × 10 −3 m ) ln ⎛
⎝ 0.800 ⎠
⎝ a⎠
Additional Problems
P26.46
Imagine the center plate is split along its midplane and pulled apart.
We have two capacitors in parallel, supporting the same ∆V and
∈ A
carrying total charge Q. The upper has capacitance C1 = 0
and the
d
∈ A
lower C2 = 0 . Charge flows from ground onto each of the outside
2d
plates so that Q1 + Q2 = Q
∆V1 = ∆V2 = ∆V
Q1 Q2 Q1d Q2 2d
=
=
=
C1 C2 ∈0 A ∈0 A
Then
(a)
(b)
Q
Q
. On the lower plate the charge is − .
3
3
Q1 =
2Q
2Q
. On the upper plate the charge is −
.
3
3
2Qd
Q1
=
3∈ 0 A
C1
2d
FIG. P26.46
Q1 = 2Q2
Q2 =
∆V =
d
2Q2 + Q2 = Q
90
P26.47
Chapter 26
(a)
Each face of P2 carries charge, so the three-plate system is equivalent to
P2
P3
P2
P1
Each capacitor by itself has capacitance
C=
−12
2
−4
2
κ ∈ 0 A 1( 8.85 × 10 C ) 7.5 × 10 m
=
= 5.58 pF
d
N ⋅ m 2 1.19 × 10 −3 m
Then equivalent capacitance = 5.58 + 5.58 = 11.2 pF .
(b)
Q = C ∆V + C ∆V = 11.2 × 10 −12 F (12 V ) = 134 pC
(c)
Now P3 has charge on two surfaces and in effect three capacitors are in parallel:
C = 3 ( 5.58 pF ) = 16.7 pF
(d)
P26.48
Only one face of P4 carries charge: Q = C ∆V = 5.58 × 10 −12 F (12 V ) = 66.9 pC
From the Example about a cylindrical capacitor,
Vb − Va = −2 ke λ ln
b
a
Vb − 345 kV = −2 (8.99 × 10 9 Nm 2 C2 ) (1.40 × 10 −6 C m ) ln
12 m
0.024 m
= −2 (8.99 ) (1.4 × 10 3 J C ) ln 500 = −1.564 3 × 10 5 V
Vb = 3.45 × 10 5 V − 1.56 × 10 5 V = 1.89 × 10 5 V
P26.49
(a)
We use the equation U = Q2/2C to find the potential energy of the capacitor. As we will
see, the potential difference ∆V changes as the dielectric is withdrawn. The initial and
1 ⎛ Q2 ⎞
1 ⎛ Q2 ⎞
final energies are U i = ⎜ ⎟ and U f = ⎜ ⎟ .
2 ⎝ Ci ⎠
2 ⎝ Cf ⎠
1 ⎛ Q2 ⎞
But the initial capacitance (with the dielectric) is Ci = κ C f . Therefore, U f = κ ⎜ ⎟ .
2 ⎝ Ci ⎠
Since the work done by the external force in removing the dielectric equals the change in
1 ⎛ Q2 ⎞ 1 ⎛ Q2 ⎞ 1 ⎛ Q2 ⎞
potential energy, we have W = U f − U i = κ ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟ (κ − 1) .
2 ⎝ Ci ⎠ 2 ⎝ Ci ⎠ 2 ⎝ Ci ⎠
To express this relation in terms of potential difference ∆Vi , we substitute Q = Ci ( ∆Vi ) , and
1
1
2
2
evaluate: W = Ci ( ∆Vi ) (κ − 1) = ( 2.00 × 10 −9 F ) (100 V) ( 5.00 − 1.00 ) = 4.00 × 10 −5 J .
2
2
The positive result confirms that the final energy of the capacitor is greater than the initial
energy. The extra energy comes from the work done on the system by the external force
that pulled out the dielectric.
(b)
The final potential difference across the capacitor is ∆V f =
Q
.
Cf
Ci
and Q = Ci ( ∆Vi ) gives ∆V f = κ ∆Vi = 5.00 (100 V ) = 500 V .
κ
Even though the capacitor is isolated and its charge remains constant, the potential
difference across the plates does increase in this case.
Substituting C f =
Capacitance and Dielectrics
*P26.50 (a)
50 m F
20 m F
a
b
30 m F
C1
FIG. P26.50a
(b )
1
= 20 µ F
1 / 30 + 1 / (20 + C1 )
gives
1
1
1
1
1
+
=
so
=
20 + C1 60
30 20 + C1 20
and C1 = 40.0 µ F
6V
(c)
50 m F
30 m F
60 m F
FIG. P26.50c
For the 50 µF, ∆V = 6.00 V
and Q = C∆V = 300 µ C .
For the 30 µF and 60 µF, the equivalent capacitance is 20 µF and Q = C∆V =
20 µF 6 V = 120 µ C .
For 30 µF, ∆V = Q/C = 120 µC/30 µF = 4.00 V .
For 20 µF and 40 µF, ∆V = 120 µC/60 µF = 2.00 V .
For 20 µF, Q = C∆V = 20 µF 2 V = 40.0 µ C .
For 40 µF, Q = C∆V = 40 µF 2 V = 80.0 µ C .
P26.51
κ = 3.00 , Emax = 2.00 × 108 V m =
For
∆Vmax
d
C=
κ ∈0 A
= 0.250 × 10 −6 F
d
A=
( 0.250 × 10 −6 ) ( 4 000 ) = 0.188 m 2
C ∆Vmax
Cd
=
=
κ ∈0 κ ∈0 Emax 3.00 (8.85 × 10 −12 ) ( 2.00 × 108 )
91
92
Chapter 26
*P26.52 (a)
The partially filled capacitor constitutes two capacitors in series, with separate capacitances
1 ∈0 A
6.5 ∈0 A
and
d (1 − f )
fd
and equivalent capacitance
6.5 ∈0 A
∈ A
1
6.5
=
= 0
= 25.0 µ F (1 − 0.846 f )−1
d (1 − f )
fd
−
+
6
.
5
d
6
.
5
df
fd
d
6
.
5
−
5
.
5
f
+
∈0 A
6.5 ∈0 A
(b)
For f = 0, the capacitor is empty so we can expect capacitance 25.0 µ F , and the
expression in part (a) agrees with this.
(c)
For f = 1 we expect 6.5(25.0 µ F) = 162 µ F. The expression in (a) becomes
25.0 µ F(1 – 0.846)–1 = 162 µ F, in agreement .
(d)
The charge on the lower plate creates an electric field in the liquid given by
E=
Q
300 µ C
=
Aκ ∈0 A 6.5 ∈0
The charge on the upper plate creates an electric field in the vacuum according to
E=
300 µ C
Q
=
A ∈0
A ∈0
The change in strength of the field at the upper surface of the liquid is described by
300 µ C Qinduced 300 µ C
+
=
A 6.5 ∈0
A ∈0
A ∈0
which gives Qinduced = 254 µ C, independent of f . The induced charge
will be opposite in sign to the charge on the top capacitor plate and
the same in sign as the charge on the lower plate .
P26.53
(a)
(b)
Put charge Q on the sphere of radius a and –Q on the other sphere. Relative to V = 0 at
infinity,
the potential at the surface of a is
Va =
keQ keQ
−
a
d
and the potential of b is
Vb =
− keQ keQ
.
+
b
d
The difference in potential is
Va − Vb =
and
C=
As d → ∞,
C=
keQ keQ keQ keQ
+
−
−
a
b
d
d
⎛
⎞
4π ∈0
Q
= ⎜
Va − Vb
⎝ (1 a ) + (1 b ) − ( 2 d ) ⎟⎠
1
1
becomes negligible compared to . Then,
a
d
1
1
1
4π ∈0
and
=
+
C
4π ∈0 a 4π ∈0 b
1 a +1 b
as for two spheres in series.
Capacitance and Dielectrics
P26.54
93
The initial charge on the larger capacitor is
Q = C ∆V = 10 µ F (15 V ) = 150 µ C
An additional charge q is pushed through the 50-V battery, giving the smaller capacitor charge q
and the larger charge 150 µ C + q .
50 V =
Then
q
150 µ C + q
+
5 µF
10 µ F
500 µ C = 2q + 150 µ C + q
q = 117 µ C
P26.55
So across the 5- µ F capacitor
∆V =
q 117 µ C
=
= 23.3 V
C
5 µF
Across the 10- µ F capacitor
∆V =
150 µ C + 117 µ C
= 26.7 V
10 µ F
3
1.00 gal
⎞ ⎛ 1.00 m ⎞ = 5.24 × 10 7 J kg
Btu gal ) (1 054 J Btu ) ⎛
−
3
3
⎝ 3.786 × 10 m ⎠ ⎜⎝ 670 kg ⎟⎠
Gasoline:
(126 000
Battery:
(12.0 J C)(100
C s)( 3 600 s)
16.0 kg
Capacitor:
1
2
( 0.100 F ) (12.0 V )2
0.100 kg
= 2.70 × 10 5 J kg
= 72.0 J kg
Gasoline has 194 times the specific energy content of the battery and 727 000 times thhat of the
capacitor.
*P26.56 (a)
The portion of the device containing the dielectric has plate area x and
capacitance C1 =
κ ∈0 x
. The unfilled part has area ( − x ) and
d
capacitance C2 =
∈
∈0 ( − x )
. The total capacitance is C1 + C2 = 0 ⎡⎣ 2 + x (κ − 1) ⎤⎦ .
d
d
1 Q2
Q2d
.
=
2
2 C
2 ∈ 0 ( + x (κ − 1))
(b)
The stored energy is U =
(c)
dU ⎞ ˆ
Q 2 d (κ − 1)
ˆ
F = −⎛
i=
2 i . When x = 0, the original value of the force
⎝ dx ⎠
2∈0 ( 2 + x (κ − 1))
is
Q 2 d (κ − 1)
î. As the dielectric slides in, the charges on the plates redistribute themselves.
2∈0 3
The force decreases to its final value
Q 2 d (κ − 1)
2 ∈ 0 3κ 2
î .
continued on next page
94
Chapter 26
(d)
2Q 2 d (κ − 1) ˆ
,F=
i.
2
2
∈ 0 3 (κ + 1)
For the constant charge on the capacitor and the initial voltage we have the relationship
At x =
Q = C0 ∆ V =
∈0 2 ∆ V
d
2 ∈ 0 ( ∆V )2 (κ − 1)
ˆi
Then the force is F =
2
d (κ + 1)
2 (8.85 × 10 −12 C2 ) 0.05 m ( 2 000 Nm )2 ( 4.5 − 1)
ˆi = 205µ N ˆi
F=
2
Nm 2 ( 0.002 m ) C2 ( 4.5 + 1)
*P26.57 The portion of the capacitor nearly filled by metal has
κ ∈ 0 ( x )
→∞
d
Q2
→0
2C
capacitance
and stored energy
The unfilled portion has
∈0 ( − x )
d
capacitance
Q=
The charge on this portion is
(a)
( − x ) Q0
The stored energy is
Q 2 ( − x ) d
Q 2 ⎡⎣( − x ) Q0 ⎤⎦
U=
=
= 0
2C 2 ∈ 0 ( − x ) d
2 ∈0 3
2
(b)
F =−
F=
2
dU
d ⎛ Q ( − x ) d ⎞
Q2d
= − ⎜⎜ 0
⎟⎟ = + 0 3
3
2 ∈0 dx
dx ⎝ 2 ∈ 0 ⎠
Q02 d
to the right (into the capacitor)
2 ∈0 3
(c)
Stress =
(d)
u=
F
Q02
=
2 ∈0 4
d
2
2
Q02
1
1 ⎛σ⎞
1 ⎛ Q ⎞
. The answers to parts (c) and (d) are
∈0 E 2 = ∈0 ⎜ ⎟ = ∈0 ⎜ 0 2 ⎟ =
2∈0 4
2
2 ⎝ ∈0 ⎠
2 ⎝ ∈0 ⎠
precisely the same.
*P26.58 One capacitor cannot be used by itself—it would burn out. She can
use two capacitors in parallel, connected in series to another two
capacitors in parallel. One capacitor will be left over. The equivalent
capacitance is
1
= 100 µ F . When 90 V is
−1
µ
F
+
200
(
) ( 200 µF )−1
FIG. P26.58
connected across the combination, only 45 V appears across each individual capacitor.
Capacitance and Dielectrics
P26.59
95
Call the unknown capacitance Cu
(
Q = Cu ( ∆Vi ) = ( Cu + C ) ∆V f
(
C ∆V f
Cu =
)
( ∆Vi ) − ( ∆Vf )
=
)
(100.0 µF ) (30.0 V ) =
(100 V − 30.0 V )
4.29 µ F
*P26.60 Consider a strip of width dx and length W at position x from the front left corner. The
κ ∈ W dx
capacitance of the lower portion of this strip is 1 0
. The capacitance of the upper portion
t x/L
κ 2 ∈0 W dx
is
. The series combination of the two elements has capacitance
t (1 − x / L )
κ κ ∈ WL dx
1
= 1 2 0
tx
t ( L − x)
κ
2 tx + κ 1tL − κ 1tx
+
κ 1 ∈0 WL dx κ 2 ∈0 WL dx
The whole capacitance is a combination of elements in parallel:
C=∫
L
0
=
L κ κ ∈ WL (κ − κ ) tdx
κ 1κ 2 ∈0 WL dx
1
1 2
0
2
1
=
∫
0
(κ 2 − κ 1 ) tx + κ 1tL (κ 2 − κ 1 ) t
(κ 2 − κ 1 ) tx + κ 1tL
L
κ 1κ 2 ∈0 WL
κ κ ∈ WL ⎡ (κ 2 − κ 1 ) tL + κ 1tL ⎤
ln
ln ⎡⎣(κ 2 − κ 1 ) tx + κ 1tL ⎤⎦ 0 = 1 2 0
⎥
0 + κ 1tL
(κ 2 − κ 1 ) t
(κ 2 − κ 1 ) t ⎢⎣
⎦
⎡⎛ κ ⎞ −1 ⎤ κ κ ∈ WL ⎡ κ ⎤
κ 1κ 2 ∈0 WL
κ 1κ 2 ∈0 WL ⎡κ 2 ⎤
=
ln
ln ⎢ 2 ⎥ = 1 2 0
ln 1
=
(κ 2 − κ 1 ) t ⎢⎣ κ 1 ⎥⎦ (−1) (κ 2 − κ 1 ) t ⎢⎣⎜⎝ κ 1 ⎟⎠ ⎥⎦ (κ 1 − κ 2 ) t ⎢⎣κ 2 ⎥⎦
(b)
The capacitor physically has the same capacitance if it is turned upside down, so the answer
should be the same with κ1 and κ2 interchanged. We have proved that it has this property in
the solution to part (a).
(c)
Let κ1 = κ2 (1 + x). Then C =
κ 2 (1 + x )κ 2 ∈0 WL
ln [1 + x ].
κ 2 xt
As x approaches zero we have C =
P26.61
(a)
C0 =
κ (1 + 0) ∈0 WL
κ ∈0 WL
as was to be shown.
x=
xt
t
∈0 A Q0
=
d
∆V0
When the dielectric is inserted at constant voltage,
Q
C = κ C0 =
∆V0
U0 =
C0 ( ∆V0 )
2
2
C ( ∆V0 ) κ C0 ( ∆V02 )
=
U=
2
2
U
=κ
U0
2
and
(b)
The extra energy comes from (part of the) electrical work done by the battery in separating
the extra charge.
Q0 = C0 ∆V0
and
Q = C ∆V0 = κ C0 ∆V0
so the charge increases according to
Q
=κ .
Q0
96
P26.62
Chapter 26
Assume a potential difference across a and b, and notice that the potential difference across the
8.00 µ F capacitor must be zero by symmetry. Then the equivalent capacitance can be determined from the following circuit:
FIG. P26.62
Cab = 3.00 µ F
P26.63
Initially (capacitors charged in parallel),
q1 = C1 ( ∆V ) = ( 6.00 µ F ) ( 250 V ) = 1 500 µ C
q2 = C2 ( ∆V ) = ( 2.00 µ F ) ( 250 V ) = 500 µ C
After reconnection (positive plate to negative plate),
qtotal
′ = q1 − q2 = 1 000 µ C
and
∆V ′ =
qtotal
′
1 000 µ C
=
= 125 V
Ctotal 8.00 µ F
Therefore,
q1′ = C1 ( ∆V ′ ) = ( 6.00 µ F ) (125 V ) = 750 µ C
q2′ = C2 ( ∆V ′ ) = ( 2.00 µ F ) (125 V ) = 250 µ C
*P26.64 Let charge λ per length be on one wire and –λ be on the other. The electric field due to the charge
on the positive wire is perpendicular to the wire, radial, and of magnitude
E+ =
λ
2π ∈0 r
The potential difference between the surfaces of the wires due to the presence of this charge is
+ wire
∆V1 = −
∫
E ⋅dr = −
− wire
λ
2π ∈0
r
dr
D − r⎞
λ
ln ⎛
=
⎝
r
r ⎠
2
π
∈
0
D−r
∫
The presence of the linear charge density − λ on the negative wire makes an identical contribution
to the potential difference between the wires. Therefore, the total potential difference is
∆V = 2 ( ∆V1 ) =
λ
D − r⎞
ln ⎛
π ∈0 ⎝ r ⎠
With D much larger than r we have nearly ∆V =
λ
D
ln ⎛ ⎞
⎝
π ∈0
r⎠
and the capacitance of this system of two wires, each of length , is
C=
π ∈0 λ
λ
Q
=
=
=
∆V ∆V ( λ π ∈0 ) ln [ D r ] ln [ D r ]
The capacitance per unit length is
π ∈0
C
=
ln [ D / r ]
Capacitance and Dielectrics
P26.65
97
By symmetry, the potential difference across 3C is zero, so the circuit reduces to
FIG. P26.65
1
1 ⎞
Ceq = ⎛
+
⎝ 2C 4C ⎠
−1
8
4
= C=
C
6
3
P26.66 The condition that we are testing is that the capacitance increases by less than 10%, or,
C′
< 1.10
C
Substituting the expressions for C and C ′ from Example 26.1, we have
ln ( b / a )
C ′ 2 ke ln ( b / 1.10 a )
=
< 1.10
=
ln ( b / 1.10 a )
C
2 ke ln ( b / a )
This becomes
b
b
b ⎞
b
1 ⎞
ln ⎛ ⎞ < 1.10 ln ⎛
= 1.10 ln ⎛ ⎞ + 1.10 ln ⎛
= 1.10 ln ⎛ ⎞ − 1.10 ln (1.10 )
⎝ a⎠
⎝ 1.10 a ⎠
⎝ a⎠
⎝ 1.10 ⎠
⎝ a⎠
We can rewrite this as,
b
−0.10 ln ⎛ ⎞ < −1.10 ln (1.10 )
⎝ a⎠
b
11.0
ln ⎛ ⎞ > 11.0 ln (1.10 ) = ln (1.10 )
⎝ a⎠
where we have reversed the direction of the inequality because we multiplied the whole expression by –1 to remove the negative signs. Comparing the arguments of the logarithms on both
sides of the inequality, we see that
b
11.0
> (1.10 ) = 2.85
a
Thus, if b > 2.85a, the increase in capacitance is less than 10% and it is more effective to increase .
98
Chapter 26
ANSWERS TO EVEN PROBLEMS
P26.2
(a) 1.00 µ F
P26.4
11.1 nF;
P26.6
( 2 N − 1) ∈0 (π − θ ) R 2
(b) 100 V
26.6 C
d
P26.8
mgd tan θ
q
P26.10
Yes; its total energy is sufficient to make the trip; 1.00 × 106 m/s.
P26.12
(a) 17.0 µ F
P26.14
(a) 2C
P26.16
Cp
2
(c) 45.0 µ C and 108 µ C
(b) 9.00 V
(b) Q1 > Q3 > Q2 (c) ∆V1 > ∆V2 = ∆V3
C p2
+
− C p Cs and
4
Cp
2
C p2
−
4
(d) Q1 and Q3 increase, Q2 decreases
− C p Cs
P26.18
(a) 398 µ F in series
P26.20
19.8 µ C
P26.22
(
P26.24
83.6 µ C
P26.26
4.47 kV
P26.28
(a) See the solution. Stored energy = 0.150 J
difference = 268 V
P26.30
(a) 400 µ C
P26.32
(a) C(∆V)2 (b) 4∆V/3 (c) 4C(∆V)2/3 (d) Positive work is done by the agent pulling the
plates apart.
R2Q
R1Q
(a) q1 =
and q2 =
(b) See the solution.
R1 + R2
R1 + R2
P26.34
(b) 2.20 µ F in parallel
) C2
3 −1
0
(b) See the solution. Potential
(b) 2.50 kN m
P26.36
(a) 13.3 nC
(b) 272 nC
P26.38
~10 −6 F and ~10 2 V for two 40 cm by 100 cm sheets of aluminum foil sandwiching a thin sheet
of plastic.
P26.40
(a) 369 pC
P26.42
(a) −9.10 ˆi + 8.40 ˆj pC ⋅ m
P26.44
See the solution.
P26.46
(a) −2Q/3 on upper plate, −Q/3 on lower plate
P26.48
189 kV
(b) 118 pF, 3.12 V (c) –45.5 nJ
(
)
(b) −20.9 nN ⋅ mkˆ
(c) 112 nJ
(b) 2Qd/3∈0A
(d) 228 nJ
Capacitance and Dielectrics
P26.50
(a) See the solution. (b) 40.0 µF (c) 6.00 V across 50 µF with charge 300 µF; 4.00 V across
30 µF with charge 120 µF; 2.00 V across 20 µF with charge 40 µF; 2.00 V across 40 µF with
charge 80 µF
P26.52
(a) 25.0 µF (1 − 0.846 f )−1 (b) 25.0 µF; the general expression agrees. (c) 162 µF; the
general expression agrees. (d) It has the same sign as the lower capacitor plate and its
magnitude is 254 µC, independent of f.
P26.54
23.3 V; 26.7 V
P26.56
∈0 ⎡⎣ 2 + x (κ − 1) ⎤⎦
d
(d) 205 µN right
(a)
(b)
Q2d
2∈0 ⎡⎣ 2 + x (κ − 1) ⎤⎦
(c)
Q 2 d (κ − 1)
2∈0 ⎡⎣ 2 + x (κ − 1) ⎤⎦
2
99
to the right
P26.58
One capacitor cannot be used by itself—it would burn out. She can use two capacitors in parallel,
connected in series to another two capacitors in parallel. One capacitor will be left over. Each of
the four capacitors will be exposed to a maximum voltage of 45 V.
P26.60
(a)
P26.62
3.00 µ F
P26.64
See the solution.
P26.66
See the solution.
κ 1κ 2 ∈0 WL κ 1
ln
(κ 1 − κ 2 ) t κ 2
(b) The capacitance is the same if k 1 and k 2 are interchanged, as it should be.
27
Current and Resistance
CHAPTER OUTLINE
27.1
27.2
27.3
27.4
27.5
27.6
Electric Current
Resistance
A Model for Electrical Conduction
Resistance and Temperature
Superconductors
Electrical Power
ANSWERS TO QUESTIONS
Q27.1
Voltage is a measure of potential difference, not of
current. “Surge” implies a flow—and only charge, in
coulombs, can flow through a system. It would also be
correct to say that the victim carried a certain current, in
amperes.
Q27.2
Geometry and resistivity. In turn, the resistivity of the
material depends on the temperature.
*Q27.3 (i) We require rL /AA = 3rL /AB. Then AA /AB = 1/3,
answer (f ).
(ii) πrA2/πrB2 = 1/3 gives rA /rB = 1/ 3, answer (e).
*Q27.4 Originally, R =
ρ
ρ ( / 3) ρ R
=
= .
. Finally, R f =
A
3A
9A 9
Answer (b).
Q27.5
The conductor does not follow Ohm’s law, and must have a resistivity that is current-dependent,
or more likely temperature-dependent.
Q27.6
The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons
more efficiently.
Q27.7
(i) The current density increases, so the drift speed must increase. Answer (a).
(ii) Answer (a).
Q27.8
The resistance of copper increases with temperature, while the resistance of silicon decreases
with increasing temperature. The conduction electrons are scattered more by vibrating atoms
when copper heats up. Silicon’s charge carrier density increases as temperature increases and
more atomic electrons are promoted to become conduction electrons.
*Q27.9 In a normal metal, suppose that we could proceed to a limit of zero resistance by lengthening the
average time between collisions. The classical model of conduction then suggests that a constant
applied voltage would cause constant acceleration of the free electrons. The drift speed and the
current would increase steadily in time.
It is not the situation envisioned in the question, but we can actually switch to zero resistance
by substituting a superconducting wire for the normal metal. In this case, the drift velocity of
electrons is established by vibrations of atoms in the crystal lattice; the maximum current is
limited; and it becomes impossible to establish a potential difference across the superconductor.
Q27.10 Because there are so many electrons in a conductor (approximately 1028 electrons/m3) the
average velocity of charges is very slow. When you connect a wire to a potential difference, you
establish an electric field everywhere in the wire nearly instantaneously, to make electrons start
drifting everywhere all at once.
101
102
Chapter 27
*Q27.11 Action (a) makes the current three times larger.
(b) causes no change in current.
(c) corresponds to a current 3 times larger.
(d) R is 1/4 as large, so current is 4 times larger.
(e) R is 2 times larger, so current is half as large.
(f) R increases by a small percentage as current has a small decrease.
(g) Current decreases by a large factor.
The ranking is then d > a > c > b > f > e > g.
*Q27.12 RA =
ρ LA
ρ 2 LB
1 ρ LB
R
=
=
= B
π ( d A / 2 )2 π ( 2 d B / 2 )2 2 π ( d B / 2 )2
2
PA = I A ∆V = (∆V )2 /RA = 2(∆V )2 /RB = 2PB Answer (e).
*Q27.13 RA =
ρA L 2 ρB L
=
= 2 RB
A
A
PA = I A ∆V = (∆V )2 /RA = (∆V )2 / 2 RB = PB / 2 Answer (f ).
*Q27.14 (i) Bulb (a) must have higher resistance so that it will carry less current and have lower power.
(ii) Bulb (b) carries more current.
*Q27.15 One ampere–hour is (1 C/s)(3 600 s) = 3 600 coulombs. The ampere–hour rating is the quantity
of charge that the battery can lift though its nominal potential difference. Answer (d).
Q27.16 Choose the voltage of the power supply you will use to drive the heater. Next calculate the
∆V 2
. Knowing the resistivity ρ of the material, choose a combination
required resistance R as
P
⎛ R⎞
of wire length and cross-sectional area to make ⎛ ⎞ = ⎜ ⎟ . You will have to pay for less
⎝ A⎠ ⎝ ρ ⎠
material if you make both and A smaller, but if you go too far the wire will have too little
surface area to radiate away the energy; then the resistor will melt.
SOLUTIONS TO PROBLEMS
Section 27.1
P27.1
Electric Current
I=
∆Q
∆t
N=
1.20 × 10 −3 C
Q
=
= 7.50 × 1015 electrons
e 1.60 × 10 −19 C electron
)
∆Q = I ∆t = ( 30.0 × 10 −6 A ( 40.0 s ) = 1.20 × 10 −3 C
Current and Resistance
P27.2
103
The molar mass of silver = 107.9 g mole and the volume V is
V = ( area ) ( thickness ) = ( 700 × 10 −4 m 2 ) ( 0.133 × 10 −3 m ) = 9.31 × 10 −6 m 3
)
)
The mass of silver deposited is mAg = ρV = (10.5 × 10 3 kg m 3 ( 9.31 × 10 −6 m 3 = 9.78 × 10 −2 kg.
And the number of silver atoms deposited is
⎛ 6.02 × 10 23 atoms ⎞ ⎛ 1 000 g ⎞
23
N = ( 9.78 × 10 −2 kg ⎜
⎟⎠ ⎜⎝ 1 kg ⎟⎠ = 5.45 × 10 atoms
107.9 g
⎝
)
∆V 12.0 V
=
= 6.67 A = 6.67 C s
R 1.80 Ω
23
−19
∆Q Ne ( 5.45 × 10 (1.60 × 10 C
∆t =
=
=
= 1.31 × 10 4 s = 3.64 h
I
I
6.67 C s
I=
)
t
P27.3
Q ( t ) = ∫ Idt = I 0τ (1 − e− t τ
0
P27.4
)
)
(a)
Q (τ ) = I 0τ (1 − e−1 =
(b)
Q (10τ ) = I 0τ (1 − e−10 =
(c)
Q ( ∞) = I 0τ (1 − e−∞ = I 0τ
( 0.632 ) I 0τ
) ( 0.999 95 ) I τ
0
)
The period of revolution for the sphere is T =
revolving charge is I =
P27.5
)
q
qω
=
.
2π
T
2π
, and the average current represented by this
ω
q = 4 t 3 + 5t + 6
2
⎛ 1.00 m ⎞
= 2.00 × 10 −4 m 2
A = ( 2.00 cm 2 ⎜
⎝ 100 cm ⎟⎠
)
P27.6
(a)
I (1.00 s ) =
(b)
J=
I=
dq
dt
dq
= (12t 2 + 5
dt t =1.00 s
)
t =1.00 s
= 17.0 A
17.0 A
I
=
= 85.0 kA m 2
A 2.00 × 10 −4 m 2
1 240 s
⎛ 120π t ⎞
∫ (100 A ) sin ⎜⎝ s ⎟⎠ dt
q = ∫ dq = ∫ Idt =
0
−100 C ⎡
⎤ +100 C
⎛π⎞
q=
cos ⎜ ⎟ − cos 0 ⎥ =
= 0.265 C
⎢
⎝ 2⎠
120π ⎣
120π
⎦
P27.7
I
8.00 × 10 −6 A
2
=
2 = 2.55 A m
A π (1.00 × 10 −3 m )
(a)
J=
(b)
From J = nevd , we have
(c)
From I =
∆Q
, we have
∆t
n=
J
2.55 A m 2
=
= 5.31 × 1010 m −3 .
evd (1.60 × 10 −19 C ) ( 3.00 × 108 m s )
∆t =
23
−19
∆Q N A e ( 6.02 × 10 ) (1.60 × 10 C )
=
=
= 1.20 × 1010 s .
8.00 × 10 −6 A
I
I
(This is about 382 years!)
104
*P27.8
Chapter 27
(a)
(b)
5.00 A
I
=
= 99.5 kA m 2
A π ( 4.00 × 10 −3 m 2
Current is the same and current density is smaller. Then I = 5.00 A ,
1
1
J 2 = J1 = 9.95 × 10 4 A/m 2 = 2.49 × 10 4 A/m 2
4
4
J=
)
A2 = 4 A1
P27.9
(a)
or
π r22 = 4 π r12
so
r2 = 2 r1 = 0.800 cm
The speed of each deuteron is given by
( 2.00 × 106 (1.60 × 10−19 J = 12 ( 2 × 1.67 × 10−27 kg v2 and
The time between deuterons passing a stationary point is t in
)
)
)
10.0 × 10 −6 C s = 1.60 × 10 −19 C t or
1
mv 2
2
v = 1.38 × 10 7 m s
q
I=
t
t = 1.60 × 10 −14 s
K=
)
)
So the distance between them is vt = (1.38 × 10 7 m s (1.60 × 10 −14 s = 2.21 × 10 −7 m .
(b)
One nucleus will put its nearest neighbor at potential
V=
9
2
2
−19
ke q (8.99 × 10 N ⋅ m C ) (1.60 × 10 C )
=
= 6.49 × 10 −3 V
2.21 × 10 −7 m
r
This is very small compared to the 2 MV accelerating potential, so repulsion within the
beam is a small effect.
P27.10
We use I = nqAvd n is the number of charge carriers per unit volume, and is identical to the
number of atoms per unit volume. We assume a contribution of 1 free electron per atom in the
relationship above. For aluminum, which has a molar mass of 27, we know that Avogadro’s
number of atoms, N A, has a mass of 27.0 g. Thus, the mass per atom is
27.0 g
27.0 g
=
= 4.49 × 10 −23 g atom
NA
6.02 × 10 23
n=
Thus,
density of aluminum
2.70 g cm 3
=
4.49 × 10 −23 g atom
mass per atom
n = 6.02 × 10 22 atoms cm 3 = 6.02 × 10 28 atoms m 3
vd =
or,
vd = 0.130 mm s
Section 27.2
P27.11
I
5.00 A
=
= 1.30 × 10 −4 m s
nqA ( 6.02 × 10 28 m −3 (1.60 × 10 −19 C ( 4.00 × 10 −6 m 2
Therefore,
)
)
)
Resistance
∆V = IR
and
R=
ρ
:
A
∆V =
I ρ
:
A
2
2 ⎛ 1.00 m ⎞
A = ( 0.600 mm ) ⎜
= 6.00 × 10 −7 m 2
⎝ 1 000 mm ⎟⎠
I=
)
−7
2
∆VA ( 0.900 V) ( 6.00 × 10 m
=
ρ
( 5.60 × 10−8 Ω ⋅ m (1.50 m )
I = 6.43 A
)
Current and Resistance
P27.12
I=
P27.13
(a)
(b)
∆V 120 V
=
= 0.500 A = 500 mA
R
240 Ω
Given
M = ρdV = ρd A where
we obtain:
A=
Thus,
=
V=
M
,
ρd
M
ρd MR
=
ρr ρd
π r 2 =
or
r=
Thus,
rd ≡ mass density,
R=
Taking rr ≡ resistivity,
(1.00 × 10 ) ( 0.500 )
(1.70 × 10 ) ( 8.92 × 10 )
−3
−8
3
P27.15
(a)
M
1.00 × 10 −3
=
π ρd π ( 8.92 × 10 3 (1.82 )
)
13
−1
ρ 4 ρ 4 (10 Ω ⋅ m ) (10 m )
=
~
= ~1018 Ω
2
−3
A π d2
π (10 m )
(b)
R=
−8
−3
4 ρ 4 (1.7 × 10 Ω ⋅ m (10 m
~
2
π d2
π ( 2 × 10 −2 m
(c)
I=
∆V 10 2 V
~
~10 −16 A
R 1018 Ω
I~
10 2 V
~10 9 A
10 −7 Ω
J = σE
*P27.16 (a)
= 1.82 m
r = 1.40 × 10 −4 m
diameter = 280 µ m
Suppose the rubber is 10 cm long and 1 mm in diameter.
R=
Section 27.3
ρr ρr ρ ρ 2
=
= r d
A
M ρd M
M
ρd
The diameter is twice this distance:
P27.14
105
)
so
σ=
)
)
~10 −7 Ω
J 6.00 × 10 −13 A m 2
−1
=
= 6.00 × 10 −15 ( Ω ⋅ m )
E
100 V m
A Model for Electrical Conduction
The density of charge carriers n is set by the material and is unaffected .
I
so it doubles .
A
doubles .
(b)
The current density is proportional to current according to J =
(c)
For larger current density in J = nevd the drift speed vd
mσ
The time between collisions τ = 2 is unchanged as long as σ does not change due to a
nq
temperature change in the conductor.
(d)
106
P27.17
Chapter 27
ρ=
so
m
We take the density of conduction electrons from an Example in the chapter text.
nq 2τ
( 9.11 × 10−31
m
τ=
=
= 2.47 × 10 −14 s
ρnq 2 (1.70 × 10 −8 ( 8.46 × 10 28 (1.60 × 10 −19 2
)
vd =
qE
τ
m
)
)
(1.60 × 10 ) E ( 2.47 × 10 )
=
−19
gives
7.84 × 10
Therefore,
E = 0.180 V m
Section 27.4
P27.18
−4
−14
9.11 × 10 −31
Resistance and Temperature
)
R = R0 ⎡⎣1 + α ( ∆T ) ⎤⎦ gives
140 Ω = (19.0 Ω ) ⎡⎣1 + ( 4.50 × 10 −3 °C ∆T ⎤⎦
Solving,
∆T = 1.42 × 10 3°C = T − 20.0°C
T = 1.44 × 10 3°C
And the final temperature is
P27.19
)
(a)
ρ = ρ0 ⎡⎣1 + α (T − T0 ) ⎤⎦ = ( 2.82 × 10 −8 Ω ⋅ m ) ⎡⎣1 + 3.90 × 10 −3 ( 30.0°) ⎤⎦ = 3.15 × 10 −8 Ω ⋅ m
(b)
J=
(c)
⎡ π 1.00 × 10 −4 m 2 ⎤
⎛ π d2 ⎞
) ⎥ = 49.9 mA
6
2 ⎢ (
6
.
35
10
A
m
I = JA = J ⎜
×
⎟=(
)
⎢
⎥
4
4
⎝
⎠
⎣
⎦
(d)
n=
0.200 V m
E
=
= 6.35 × 10 6 A m 2
ρ 3.15 × 10 −8 Ω ⋅ m
6.02 × 10 23 electrons
⎡26.98 g ( 2.70 × 10 6 g m 3 )⎤
⎣
⎦
vd =
(e)
= 6.02 × 10 28 electrons m 3
(6.35 × 106 A m 2 )
J
=
= 659 µ m s
ne ( 6.02 × 10 28 electrons m 3 ) (1.60 × 10 −19 C)
∆V = E = ( 0.200 V m ) ( 2.00 m ) = 0.400 V
*P27.20 We require 10 Ω =
any ∆T
10 Ω =
3.5 × 10 −5 Ω ⋅ m 1 1.5 × 10 −6 Ω ⋅ m 2
+
π (1.5 × 10 −3 m )2
π (1.5 × 10 −3 m )2
and for
3.5 × 10 −5 Ω ⋅ m 1 ⎛
1.5 × 10 −6 Ω ⋅ m 2
−3 ∆T ⎞
1
−
0
5
×
10
.
+
⎜
⎟
π (1.5 × 10 −3 m )2 ⎝
°C ⎠ π (1.5 × 10 −3 m )2
⎛
−3 ∆T ⎞
⎜⎝ 1 + 0.4 × 10
⎟
°C ⎠
simplifying gives 10 = 4.951 5 1 + 0.212 21 2
and
0 = – 2.475 7 × 10–3 1 + 8.488 3 × 10–5 2
These conditions are just sufficient to determine 1 and 2. The design goal can be met.
We have 2 = 29.167 1
and
so
10 = 4.951 5 1 + 0.212 21 (29.167 1)
1 = 10/11.141 = 0.898 m = 1
2 = 26.2 m
Current and Resistance
P27.21
R = R0 [1 + α T ]
R − R0 = R0α∆T
R − R0
= α∆T = ( 5.00 × 10 −3 25.0 = 0.125
R0
)
P27.22
For aluminum,
R=
α E = 3.90 × 10 −3°C−1
(Table 27.2)
α = 24.0 × 10 −6°C−1
(Table 19.1)
(1 + α E ∆T ) = 1.234 Ω ⎛ 1.39 ⎞ = 1.71 Ω
ρ ρ0 (1 + α E ∆T ) (1 + α∆T )
=
= R0
(
)⎜
2
A
(1 + α∆T )
⎝ 1.002 4 ⎟⎠
A (1 + α∆T )
Section 27.5
Superconductors
Problem 50 in Chapter 43 can be assigned with this section.
Section 27.6
P27.23
I=
Electrical Power
P
∆V
and R =
P27.24
=
600 W
= 5.00 A
120 V
∆V 120 V
=
= 24.0 Ω
I
5.00 A
P = I ∆V = 500 × 10 −6 A (15 × 103 V ) = 7.50 W
*P27.25 The energy that must be added to the water is
Q = mc∆T = (109 kg ) ( 4 186 J kg°C ) ( 29.0°C ) = 1.32 × 10 7 J
Thus, the power supplied by the heater is
W
Q 1.32 × 10 7 J
P= = =
= 8 820 W
∆t ∆t
25 × 60 s
and the resistance is R =
*P27.26 (a)
efficiency =
I=
( ∆V )2
P
=
( 220 V )2
8 820 W
= 5.49 Ω .
2.50 hp(746 W/1 hp)
mechanical power output
= 0.900 =
(120 V) I
total power input
1 860 J/s
2 070 J/s
=
= 17.3 A
0.9(120 V) 120 J/C
(b)
energy input = Pinput ∆t = (2 070 J/s) 3 (3 600 s) = 2.24 × 10 7 J
(c)
S/ 0.16 ⎞ ⎛ k J
h ⎞
= $ 0.995
cost = 2.24 × 107 J ⎛
⎝ 1 kWh ⎠ ⎝ 10 3 W s 3 600 s ⎠
107
108
Chapter 27
2
P27.27
P ( ∆V )2 R ⎛ ∆V ⎞ ⎛ 140 ⎞ 2
=
=
=
= 1.361
P0 ( ∆V0 )2 R ⎜⎝ ∆V0 ⎟⎠ ⎝ 120 ⎠
⎛ P − P0 ⎞
⎛P
⎞
∆% = ⎜
(100%) = ⎜ − 1⎟ (100%) = (1.3661 − 1)100% = 36.1%
⎟
P
P
⎝
⎝ 0 ⎠
⎠
0
P27.28
The battery takes in energy by electric transmission
P ∆t = ( ∆V ) I ( ∆t ) = 2.3 J C (13.5 × 10 −3 C s ) 4.2 h ⎛⎝
3 600 s ⎞
= 469 J
1h ⎠
It puts out energy by electric transmission
( ∆V ) I ( ∆t ) = 1.6 J C (18 × 10 −3 C s ) 2.4 h ⎛
3 600 s ⎞
= 249 J
⎝ 1h ⎠
useful output 249 J
=
= 0.530
total input
469 J
(a)
efficiency =
(b)
The only place for the missing energy to go is into internal energy:
469 J = 249 J + ∆ Eint
∆ Eint = 221 J
(c)
We imagine toasting the battery over a fire with 221 J of heat input:
Q = mc∆T
Q
221 J
∆T =
=
mc 0.015 kg
P27.29
P = I ( ∆V ) =
R
(110 V) = 24.2 Ω
(500 W)
2
= 500 W
R=
−4
RA ( 24.2 Ω ) π ( 2.50 × 10 m )
=
=
= 3.17 m
1.50 × 10 −6 Ω ⋅ m
ρ
2
(a)
ρ
R= A
(b)
R = R0 [1 + α∆T ] = 24.2 Ω ⎡⎣1 + ( 0.400 × 10 −3 ) (1180 )⎤⎦ = 35.6 Ω
P=
P27.30
( ∆V )2
kg°C
= 15.1°C
975 J
R=
( ∆V )2
R
so
=
(110 )2
35.6
= 340 W
−6
ρ (1.50 × 10 Ω ⋅ m ) 25.0 m
=
= 298 Ω
2
A
π ( 0.200 × 10 −3 m )
∆V = IR = ( 0.500 A)( 298 Ω) = 149 V
∆V 149 V
=
= 5.97 V m
25.0 m
(a)
E=
(b)
P = ( ∆V ) I = (149 V ) ( 0.500 A ) = 74.6 W
(c)
R = R0 ⎡⎣1 + α (T − T0 )⎤⎦ = 298 Ω ⎡⎣1 + ( 0.400 × 10 −3 °C) 320°C⎤⎦ = 337 Ω
I=
∆V (149 V )
=
= 0.443 A
R ( 337 Ω )
P = ( ∆V ) I = (149 V ) ( 0.443 A ) = 66.1 W
Current and Resistance
P27.31
(a)
109
⎛ 1 C ⎞ ⎛ 1 J ⎞ ⎛1 W ⋅ s ⎞
∆U = q ( ∆V ) = It ( ∆V ) = (55.0 A ⋅ h )(12.0 V) ⎜
⎟⎜
⎟⎜
⎟
⎝1 A ⋅ s ⎠ ⎝1 V ⋅ C ⎠ ⎝ 1 J ⎠
= 660 W ⋅ h = 0.660 kWh
(b)
*P27.32 (a)
⎛ $0.060 0 ⎞
Cost = 0.660 kWh ⎜
⎟ = 3.96¢
⎝ 1 kWh ⎠
The resistance of 1 m of 12-gauge copper wire is
R=
−8
ρ
ρ
4 ρ 4 (1.7 × 10 Ω ⋅ m )1 m
=
=
=
= 5.14 × 10 −3 Ω
2
A π ( d 2)
π d 2 π ( 0.205 3 × 10 −2 m )2
The rate of internal energy production is P = I ∆V = I 2 R = ( 20 A ) 5.14 × 10 −3 Ω = 2.05 W .
2
(b)
PAl = I 2 R =
I 2 4 ρAl π d2
PAl ρAl
=
PCu ρCu
PAl =
2.82 × 10 −8 Ω ⋅ m
2.05 W = 3.41 W
1.7 × 10 −8 Ω ⋅ m
Aluminum of the same diameter will get hotter than copper. It would not be as safe. If it is
surrounded by thermal insulation, it could get much hotter than a copper wire.
P27.33
The energy taken in by electric transmission for the fluorescent lamp is
P∆t = 11 J s (100 h ) ⎛⎝
3 600 s ⎞
= 3.96 × 10 6 J
1h ⎠
⎛ $0.08 ⎞ ⎛ k ⎞ ⎛ W ⋅ s ⎞ ⎛ h ⎞
= $0.088
cost = 3.96 × 10 6 J ⎜
⎝ kWh ⎟⎠ ⎜⎝ 1 000 ⎟⎠ ⎝ J ⎠ ⎜⎝ 3 600 s ⎟⎠
For the incandescent bulb,
P∆t = 40 W (100 h ) ⎛⎝
3 600 s ⎞
= 1.44 × 10 7 J
1h ⎠
⎛ $0.08 ⎞
cost = 1.44 × 10 7 J ⎜
= $0.32
⎝ 3.6 × 10 6 J ⎟⎠
saving = $0.32 − $0.088 = $0.232
P27.34
The total clock power is
( 270 × 10
From e =
6
J s ⎞ ⎛ 3 600 s ⎞
= 2.43 × 1012 J h
clocks ) ⎛ 2.50
⎝
clock ⎠ ⎝ 1 h ⎠
Wout
, the power input to the generating plants must be:
Qin
Qin Wout ∆ t 2.43 × 1012 J h
=
=
= 9.72 × 1012 J h
∆t
0.250
e
and the rate of coal consumption is
1.00 kg coal ⎞
Rate = ( 9.72 × 1012 J h ) ⎛
= 2.95 × 10 5 kg coal h = 295 metric ton h
⎝ 33.0 × 10 6 J ⎠
110
P27.35
Chapter 27
P = I ( ∆V ) = (1.70 A ) (110 V ) = 187 W
Energy used in a 24-hour day = ( 0.187 kW ) ( 24.0 h ) = 4.49 kWh.
⎛ $0.060 0 ⎞
= $0.269 = 26.9¢ .
Therefore daily cost = 4.49 kWh ⎜
⎝ kWh ⎟⎠
P27.36
P = I ∆V = ( 2.00 A ) (120 V ) = 240 W
∆Eint = ( 0.500 kg ) ( 4 186 J kg ⋅ °C ) ( 77.0°C ) = 161 kJ
∆t =
P27.37
∆Eint
P
=
1.61 × 10 5 J
= 672 s
240 W
At operating temperature,
(a)
P = I ∆V = (1.53 A ) (120 V ) = 184 W
(b)
Use the change in resistance to find the final operating temperature of the toaster.
R = R0 (1 + α∆T
∆T = 441°C
P27.38
)
(
120 120 ⎡
=
1 + 0.400 × 10 −3 ∆T ⎤⎦
1.53 1.80 ⎣
T = 20.0°C + 441°C = 461°C
You pay the electric company for energy transferred in the amount E = P ∆t .
(a)
P27.39
)
P ∆t = 40 W ( 2 weeks ) ⎛⎝
7 d ⎞ ⎛ 86 400 s ⎞ ⎛ 1 J ⎞
= 48.4 MJ
1 week ⎠ ⎝ 1 d ⎠ ⎝ 1 W ⋅ s ⎠
P ∆t = 40 W ( 2 weeks ) ⎛⎝
7 d ⎞ ⎛ 24 h ⎞ ⎛ k ⎞
= 13.4 kWh
1 week ⎠ ⎝ 1 d ⎠ ⎜⎝ 1 000 ⎟⎠
P ∆t = 40 W ( 2 weeks ) ⎛⎝
7 d ⎞ ⎛ 24 h ⎞ ⎛ k ⎞ ⎛ 0.12 $ ⎞
⎜
⎟ = $1.61
1 week ⎠ ⎝ 1 d ⎠ ⎜⎝ 1 000 ⎟⎠ ⎝ kWh ⎠
1h ⎞⎛ k ⎞
60 min ⎠ ⎜⎝ 1 000 ⎟⎠
(b)
P ∆t = 970 W ( 3 min ) ⎛⎝
(c)
P ∆t = 5 200 W ( 40 min ) ⎛⎝
⎛ 0.12 $ ⎞
⎜⎝
⎟ = $0.005 82 = 0.582¢
kWh ⎠
1h ⎞⎛ k ⎞
60 min ⎠ ⎜⎝ 1 000 ⎟⎠
⎛ 0.12 $ ⎞
⎜⎝
⎟ = $0.416
kWh ⎠
Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the
dryer is
1 kWh ⎞
P ∆t = ( 400 J s ) ( 600 s d ) ( 365 d ) ≈ 9 × 10 7 J ⎛⎝
≈ 20 kWh
6
⎠
3.6 × 10 J
We suppose that electrically transmitted energy costs on the order of ten cents per kilowatt-hour.
Then the cost of using the dryer for a year is on the order of
Cost ≈ ( 20 kWh ) ( $0.10 kWh ) = $2 ~$1
Current and Resistance
111
Additional Problems
*P27.40 (a)
I=
R=
∆V
R
( ∆V )2
P = I ∆V =
so
(120 V )
R
( ∆V ) (120 V )2
R=
=
= 144 Ω
P
100 W
2
2
(b)
= 576 Ω
and
25.0 W
P 25.0 W
Q 1.00 C
I=
=
= 0.208 A =
=
∆V
120 V
∆t
∆t
1.00 C
∆t =
= 4.80 s
0.208 A
(c)
The charge itself is the same. It comes out at a location that is at lower potential.
1.00 J
∆U 1.00 J
∆t =
= 0.040 0 s
P = 25.0 W =
=
25.0
W
∆t
∆t
P
=
( ∆V )2
The energy itself is the same. It enters the bulb by electrical transmission and leaves by
heat and electromagnetic radiation.
(d)
∆U = P∆t = ( 25.0 J s ) (86 400 s d ) ( 30.0 d ) = 64.8 × 10 6 J
The electric company sells energy .
⎛ $0.070 0 ⎞ ⎛ k ⎞ ⎛ W ⋅ s ⎞ ⎛ h ⎞
= $1.26
Cost = 64.8 × 10 6 J ⎜
⎝ kWh ⎟⎠ ⎜⎝ 1 000 ⎟⎠ ⎜⎝ J ⎟⎠ ⎜⎝ 3 600 s ⎟⎠
Cost per joule =
⎞
$0.070 0 ⎛
kWh
= $1.94 × 10 −8 J
⎜
6
kWh ⎝ 3.60 × 10 J ⎟⎠
*P27.41 The heater should put out constant power
P=
)
(
( 0.250 kg) ( 4 186 J ) (100°C − 20°C ) ⎛ 1 min ⎞
Q mc T f − Ti
=
=
= 349 J s
⎝ 60 s ⎠
∆t
∆t
kg ⋅ °C ( 4 min )
Then its resistance should be described by
P = ( ∆V ) I =
( ∆V ) ( ∆V )
R
R=
( ∆V )2 = (120
P
J C)
= 41.3 Ω
349 J s
2
Its resistivity at 100°C is given by
)
(
ρ = ρ0 ⎡⎣1 + α ( T − T0 ) ⎤⎦ = 1.50 × 10 −6 Ω ⋅ m ⎡⎣1 + 0.4 × 10 −3 (80 ) ⎤⎦ = 1.55 × 10 −6 Ω ⋅ m
Then for a wire of circular cross section
4
R=ρ =ρ 2 =ρ
A
πr
π d2
(
41.3 Ω = 1.55 × 10 −6 Ω ⋅ m
) π4d
2
d 2 = 4.77 × 10 −8 m = 2.09 × 10 +7 m
or
d2
One possible choice is = 0.900 m and d = 2.07 × 10 −4 m. If and d are made too small,
the surface area will be inadequate to transfer heat into the water fast enough to prevent
overheating of the filament. To make the volume less than 0.5 cm 3 , we want and d less
π d2
= 0.5 × 10 −6 m 3 . Substituting d 2 = 4.77 × 10 −8 m gives
than those described by
4
π
( 4.77 × 10 −8 m ) 2 = 0.5 × 10 −6 m 3 , = 3.65 m and d = 4.18 × 10 −4 m. Thus our answer is:
4
Any diameter d and length related by d 2 = 4.77 × 10 −8 m would have the right resistance.
One possibility is length 0.900 m and diameter 0.207 mm, but such a small wire might overheat
rapidly if it were not surrounded by water. The volume can be less than 0.5 cm 3 .
)
(
(
(
)
)
112
P27.42
Chapter 27
1
1 Q2
Q ∆Vi =
.
2
2 C
When the switch is closed, charge Q distributes itself over the plates of C and 3C in
parallel, presenting equivalent capacitance 4C. Then the final potential difference is
The original stored energy is U i =
(a)
∆V f =
Q
4C
for both.
Q
Q
C=
. The larger capacitor
4C
4
The smaller capacitor then carries charge C ∆V f =
(b)
carries charge 3C
Q
3Q
=
.
4C
4
( )
1
The smaller capacitor stores final energy C ∆V f
2
(c)
2
2
Q2
1 ⎛ Q⎞
=
. The larger
= C⎜
2 ⎝ 4C ⎟⎠
32C
2
capacitor possesses energy
Q2
3Q 2 Q 2
+
=
. The loss of potential energy is the energy
32C 32C 8C
3Q 2
Q2 Q2
appearing as internal energy in the resistor:
=
+ ∆ Eint
∆Eint =
8C
2C 8C
(d)
P27.43
⎛ Q⎞
1
3Q 2
=
.
3C ⎜
⎟
2 ⎝ 4C ⎠
32C
The total final energy is
We begin with the differential equation
α=
ρ
(a)
∫
Separating variables,
ρ0
⎛ ρ⎞
ln ⎜ ⎟ = α ( T − T0
⎝ ρ0 ⎠
)
and
1 dρ
ρ dT
dρ
= α dT
ρ T∫
T
0
α T −T
ρ = ρ0 e ( 0 ) .
From the series expansion e x ≈ 1 + x , ( x << 1) , we have
(b)
ρ ≈ ρ0 ⎡⎣1 + α ( T − T0 ) ⎤⎦ .
P27.44
We find the drift velocity from
vd =
v=
I = nqvd A = nqvd π r 2
I
1 000 A
−4
=
ms
2 = 2.35 × 10
2
−3
28
nqπ r
8.46 × 10 m (1.60 × 10 −19 C ) π (10 −2 m )
x
t
t=
*P27.45 From ρ =
200 × 10 3 m
x
= 8.50 × 108 s = 27.0 yr
=
v 2.35 × 10 −4 m s
)
RA ( ∆V A
=
I we compute
(m)
R (Ω)
ρ (Ω ⋅ m)
0.540
10.4
1.41 × 10 −6
1.028
21.1
1.50 × 10 −6
1.543
31.8
1.50 × 10 −6
ρ = 1.47 × 10 −6 Ω ⋅ m . With its uncertainty range from 1.41 to 1.50, this average value agrees
with the tabulated value of 1.50 × 10 −6 Ω ⋅ m in Table 27.2.
Current and Resistance
P27.46
113
2 wires → = 100 m
R=
0.108 Ω
(100 m = 0.036 0 Ω
300 m
)
(a)
( ∆V )
(b)
P = I ( ∆V ) = (110 A ) (116 V ) = 12.8 kW
(c)
Pwires = I 2 R = (110 A )2 ( 0.036 0 Ω ) = 436 W
*P27.47 (a)
( 0 − 4.00 V = 8.00 ˆi V m
dV ˆ
E=−
i=−
dx
( 0.500 − 0 m
home
= ( ∆V
)
line
)
)
− IR = 120 − (110 ( 0.036 0 = 116 V
)
)
(
)
(b)
R=
4.00 × 10 −8 Ω ⋅ m ( 0.500 m )
ρ
=
= 0.637 Ω
2
A
π 1.00 × 10 −4 m
(c)
I=
∆V
4.00 V
=
= 6.28 A
R
0.637 Ω
(d)
(e)
*P27.48 (a)
(
)
6.28 A
I
=
= 2.00 × 108 A m 2 = 200 MA m 2
2
A π (1.00 × 10 −4 m )
are both in the x direction.
J=
)(
(
The field and the current
)
ρ J = 4.00 × 10 −8 Ω ⋅ m 2.00 × 108 A m 2 = 8.00 V m = E
dV (x ) ˆ V ˆ
E=−
i=
i
dx
L
(b)
R=
ρ
4ρL
=
A
π d2
(c)
I=
∆V
Vπ d 2
=
4ρ L
R
(d)
J=
I
V
=
ρL
A
(e)
ρJ =
V
= E
L
The field and the current are both in the x direction.
114
P27.49
Chapter 27
(a)
P = I ∆V
P
so I =
(b)
∆t =
∆V
∆U
P
=
=
8.00 × 10 3 W
= 667 A
12.0 V
2.00 × 10 7 J
= 2.50 × 10 3 s
8.00 × 10 3 W
and ∆ x = v∆ t = ( 20.0 m s ) ( 2.50 × 10 3 s ) = 50.0 km
*P27.50 (a)
(b)
ρ ρ 0 ⎡⎣1 + α (T − T0 )⎦⎤ 0 ⎣⎡1 + α ′ (T − T0 )⎤⎦
=
,
A
A0 ⎡⎣1 + 2α ′ (T − T0 )⎤⎦
We begin with
R=
which reduces to
R=
For copper:
ρ0 = 1.70 × 10 −8 Ω ⋅ m , α = 3.90 × 10 −3°C−1 , and
R0 ⎡⎣1 + α (T − T0 ) ⎤⎦ ⎡⎣1 + α ′ (T − T0 ) ⎤⎦
⎡⎣1 + 2α ′ (T − T0 ) ⎤⎦
α ′ = 17.0 × 10 −6°C−1
R0 =
−8
ρ0 0 (1.70 × 10 ) ( 2.00 )
=
2 = 1.08 Ω
A0
π ( 0.100 × 10 −3 )
The simple formula for R gives:
R = (1.08 Ω) ⎡⎣1 + (3.90 × 10 −3°C−1 ) (100°C − 20.0°C)⎤⎦ = 1.420 Ω
while the more complicated formula gives:
R=
(1.08 Ω ) ⎡⎣1 + ( 3.90 × 10 −3°C−1 ) (80.0°C ) ⎤⎦ ⎡⎣1 + (17.0 × 10 −6°C−1 ) (80.0°C )
⎡⎣1 + 2 (17.0 × 10 −6°C−1 ) (80.0°C ) ⎤⎦
= 1.418 Ω
The results agree to three digits. The variation of resistance with temperature is typically a
much larger effect than thermal expansion in size.
Current and Resistance
P27.51
115
Let a be the temperature coefficient at 20.0°C, and α ′ be the temperature coefficient at 0°C.
)
Then ρ = ρ 0 ⎡⎣1 + α (T − 20.0°C)⎤⎦, and ρ = ρ ′ ⎡⎣1 + α ′ ( T − 0°C ⎤⎦ must both give the correct
resistivity at any temperature T. That is, we must have:
ρ0 [1 + α (T − 20.0°C )] = ρ ′ [1 + α ′ (T − 0°C )]
Setting T = 0 in equation (1) yields:
ρ ′ = ρ0 ⎡⎣1 − α ( 20.0°C ) ⎤⎦ ,
and setting T = 20.0°C in equation (1) gives:
ρ0 = ρ ′ [1 + α ′ ( 20.0°C )]
Put ρ ′ from the first of these results into the second to obtain:
ρ0 = ρ0 [1 − α ( 20.0°C )][1 + α ′ ( 20.0°C )]
Therefore
)
1 + α ′ ( 20.0°C =
which simplifies to
1
1 − α ( 20.0°C
)
α′ =
α
[1 − α ( 20.0°C)]
From this, the temperature coefficient, based on a reference temperature of 0°C, may be
computed for any material. For example, using this, Table 27.2 becomes at 0°C :
Material
Temp Coefficients at 0°C
Silver
4.1 × 10 −3 °C
Copper
4.2 × 10 −3 °C
Gold
3.6 × 10 −3 °C
Aluminum
4.2 × 10 −3 °C
Tungsten
4.9 × 10 −3 °C
Iron
5.6 × 10 −3 °C
Platinum
4.25 × 10 −3 °C
Lead
4.2 × 10 −3 °C
Nichrome
0.4 × 10 −3 °C
Carbon
−0.5 × 10 −3 °C
Germanium
−24 × 10 −3 °C
Silicon
−30 × 10 −3 °C
(1)
116
P27.52
Chapter 27
(a)
A thin cylindrical shell of radius r, thickness dr, and length L contributes resistance
dR =
⎛ ρ ⎞ dr
ρd ρdr
=
=⎜
A
( 2π r ) L ⎝ 2π L ⎟⎠ r
The resistance of the whole annulus is the series summation of the contributions of the thin
shells:
R=
(b)
ρ
2π L
rb
∫
ra
⎛r ⎞
dr
ρ
=
ln ⎜ b ⎟
r
2π L ⎝ ra ⎠
In this equation
⎛r ⎞
∆V
ρ
ln ⎜ b ⎟
=
I
2π L ⎝ ra ⎠
ρ=
we solve for
2π L ∆V
I ln ( rb ra )
*P27.53 The original resistance is Ri = rLi/Ai.
The new length is L = Li + d L = Li(1 + d ).
Constancy of volume implies AL = AiLi so
The new resistance is R =
A=
Ai Li
Ai Li
Ai
=
=
L
Li (1 + δ ) (1 + δ )
ρ L ρ Li (1 + δ )
=
= Ri (1 + δ )2 = Ri (1 + 2δ + δ 2 ).
A
Ai / (1 + δ )
The result is exact if the assumptions are precisely true. Our derivation contains no approximation steps where delta is assumed to be small.
P27.54
Each speaker receives 60.0 W of power. Using
I=
P
R
=
P = I 2 R, we then have
60.0 W
= 3.87 A
4.00 Ω
The system is not adequately protected since the fuse should be set to melt at 3.87 A, or lesss .
P27.55
(a)
∆V = −E ⋅ or
dV = −E ⋅ dx
∆V = − IR = −E ⋅ I=
(b)
P27.56
dV
dq E ⋅ A
A
dV
=
=
E ⋅ = E = −σ A
= σA
ρ
ρ
dx
dt
R
dx
Current flows in the direction of decreasing voltage. Energy flows by heat in the direction
of decreasing temperature.
From the geometry of the longitudinal section of the resistor shown in
the figure, we see that
(b − r ) (b − a )
y
From this, the radius at a distance y from the base is
For a disk-shaped element of volume dR =
Using the integral formula
du
∫ (au + b)
2
=−
ρdy
:
π r2
1
,
a ( au + b )
=
h
y
FIG. P27.56
r = (a − b) + b
h
h
ρ
dy
R= ∫
π 0 ⎡⎣( a − b ) ( y h ) + b ⎤⎦ 2
R=
ρ h
π ab
Current and Resistance
P27.57
ρ dx
=
A
∫
117
ρ dx
y −y
where y = y1 + 2 1 x
wy
L
R=
∫
R=
⎡
ρL
ρL
dx
y −y
=
ln ⎢ y1 + 2 1
∫
L
w 0 y1 + [( y2 − y1 ) L ] x w ( y2 − y1 ) ⎣
⎤
x⎥
⎦
L
0
FIG. P27.57
⎛y ⎞
ρL
R=
ln ⎜ 2 ⎟
w ( y2 − y1 ) ⎝ y1 ⎠
*P27.58 A spherical layer within the shell, with radius r and thickness dr, has resistance
ρ dr
4π r 2
The whole resistance is the absolute value of the quantity
dR =
r
b
rb
a
ra
R = ∫ dR = ∫
b
ρdr
ρ r −1
ρ ⎛ 1 1⎞ ρ ⎛ 1 1⎞
=
−
=
=−
− +
2
4π r
4π −1 ra
4π ⎜⎝ ra rb ⎟⎠ 4π ⎜⎝ ra rb ⎟⎠
*P27.59 Coat the surfaces of entry and exit with material of much higher conductivity than the bulk material of the object. The electric potential will be essentially uniform over each of these electrodes.
Current will be distributed over the whole area where each electrode is in contact with the resistive object.
P27.60
(a)
The resistance of the dielectric block is R =
The capacitance of the capacitor is C =
Then RC =
P27.61
ρ
d
.
=
A σA
κ∈0 A
.
d
d κ ∈0 A κ ∈0
=
is a characteristic of the material only.
σA d
σ
κ ∈0 ρ κ ∈0 75 × 1016 Ω ⋅ m (3.78) 8.85 × 10 −12 C2
= 1.79 × 1015 Ω
=
=
14 × 10 −9 F
N ⋅ m2
σC
C
(b)
R=
(a)
Think of the device as two capacitors in parallel. The one on the left has κ 1 = 1 ,
⎛
⎞
A1 = ⎜ + x ⎟ . The equivalent capacitance is
⎝2
⎠
κ 1 ∈0 A1 κ 2 ∈0 A2 ∈0 ⎛ κ ∈0 ⎛ ∈ +
=
+ x⎞ +
− x ⎞ = 0 ( + 2 x + κ − 2κ x )
2d
d
d
d ⎝2 ⎠
d ⎝2 ⎠
(b)
The charge on the capacitor is Q = C ∆V
∈0 ∆V
( + 2 x + κ − 2κ x )
2d
The current is
Q=
∈ ∆V v
dQ dQ dx ∈0 ∆V
=
=
( 0 + 2 + 0 − 2κ ) v = − 0
(κ − 1)
2d
dt
dx dt
d
The negative value indicates that the current drains charge from the capacitor. Positive
I=
current is clockwise
∈0 ∆V v
(κ − 1) .
d
118
P27.62
Chapter 27
⎡
⎛ e∆V ⎞ ⎤
∆V
I = I 0 ⎢exp ⎜
⎟ − 1⎥ and R =
I
k
T
⎝ B ⎠ ⎦
⎣
with I 0 = 1.00 × 10 −9 A, e = 1.60 × 10 −19 C, and kB = 1.38 × 10 −23 J K
The following includes a partial table of calculated values and a graph for each of the specified
temperatures.
(i)
For T = 280 K:
∆V (V) I ( A)
R (Ω)
0.400 0.015 6 25.6
0.440 0.081 8
5.338
0.480 0.429
1.12
0.520 2.25
0.232
0.560 11.8
0.047 6
0.600 61.6
0.009 7
FIG. P27.62(i)
(ii)
For T = 300 K:
∆V (V) I ( A)
R (Ω)
0.400 0.005 77.3
0.440 0.024 18.1
0.480 0.114
4.22
0.520 0.534 0.973
0.560 2.51
0.223
0.600 11.8
0.051
FIG. P27.62(ii)
(iii) For T = 320 K:
∆V ( V) I ( A )
0.400
0.440
0.480
0.520
0.560
0.600
R (Ω)
0.002 0 203
0.008 4
52.5
0.035 7
13.4
0.152
3.42
0.648
0.864
2.76
0.217
FIG. P27.62(iii)
Current and Resistance
P27.63
The volume of the gram of gold is given by ρ =
V=
m
V
10 −3 kg
m
=
= 5.18 × 10 −8 m 3 = A ( 2.40 × 10 3 m )
ρ 19.3 × 10 3 kg m 3
A = 2.16 × 10 −11 m 2
R=
P27.64
3
−8
ρ 2.44 × 10 Ω ⋅ m ( 2.4 × 10 m )
=
= 2.71 × 10 6 Ω
A
2.16 × 10 −11 m 2
⎛ 0.500 Ω ⎞
The resistance of one wire is ⎜
⎟ (100 mi) = 50.0 Ω.
⎝ mi ⎠
The whole wire is at nominal 700 kV away from ground potential, but the potential difference
between its two ends is
IR = (1 000 A ) ( 50.0 Ω ) = 50.0 kV
Then it radiates as heat power
P27.65
R = R0 ⎡⎣1 + α (T − T0 )⎤⎦
In this case, I =
I0
,
10
P = ( ∆V ) I = ( 50.0 × 103 V ) (1 000 A ) = 50.0 MW .
⎤
1⎡R
1 ⎡ I0
⎤
⎢ − 1⎥ = T0 + ⎢ − 1⎥
α ⎣ R0 ⎦
α⎣I
⎦
1
9
T = T0 + ( 9 ) = 20° +
= 2 020°C
α
0.004 50 °C
T = T0 +
so
so
ANSWERS TO EVEN PROBLEMS
P27.2
3.64 h
P27.4
qw /2p
P27.6
0.265 C
P27.8
(a) 99.5 kA/m2 (b) Current is the same, current density is smaller. 5.00 A, 24.9 kA/m2,
0.800 cm
P27.10
0.130 mm/s
P27.12
500 mA
P27.14
(a) ~1018 Ω (b) ~10−7 Ω
P27.16
(a) no change
P27.18
1.44 × 103 °C
P27.20
She can meet the design goal by choosing 1 = 0.898 m and 2 = 26.2 m.
P27.22
1.71 Ω
P27.24
7.50 W
(b) doubles
(c) ~100 aA, ~1 GA
(c) doubles
(d) no change
119
120
Chapter 27
P27.26
(a) 17.3 A (b) 22.4 MJ
(c) $0.995
P27.28
(a) 0.530
P27.30
(a) 5.97 V/m
P27.32
(a) 2.05 W (b) 3.41 W. It would not be as safe. If surrounded by thermal insulation, it would
get much hotter than a copper wire.
P27.34
295 metric ton h
P27.36
672 s
P27.38
(a) $1.61
P27.40
(a) 576 Ω and 144 Ω (b) 4.80 s. The charge itself is the same. It is at a location that is lower in
potential. (c) 0.040 0 s. The energy itself is the same. It enters the bulb by electric transmission
and leaves by heat and electromagnetic radiation. (d) $1.26, energy, 1.94 × 10−8 $/J
P27.42
(a) Q/4C (b) Q/4 and 3Q/4
P27.44
8.50 × 108 s = 27.0 yr
P27.46
(a) 116 V (b) 12.8 kW
P27.48
(a) E = V/L in the x direction
(e) See the solution.
P27.50
(a) See the solution.
P27.52
(a) R =
P27.54
No. The fuses should pass no more than 3.87 A.
P27.56
See the solution.
P27.58
See the solution.
P27.60
(b) 1.79 PΩ
P27.62
See the solution.
P27.64
50.0 MW
(b) 221 J (c) 15.1°C
(b) 74.6 W (c) 66.1 W
(b) $0.005 82
ρ
r
ln b
2π L ra
(c) $0.416
(c) Q2/32C and 3Q2/32C
(d) 3Q2/8C
(c) 436 W
(b) R = 4rL/pd 2
(c) I = Vpd 2/4rL
(b) 1.418 Ω nearly agrees with 1.420 Ω.
(b) ρ =
2π L ∆V
I ln ( rb ra )
(d) J = V/rL
28
Direct Current Circuits
CHAPTER OUTLINE
28.1
28.2
28.3
28.4
28.5
28.6
Electromotive Force
Resistors in Series and Parallel
Kirchhoff’s Rules
RC Circuits
Electrical Meters
Household Wiring and Electrical
Safety
ANSWERS TO QUESTIONS
Q28.1
No. If there is one battery in a circuit, the current inside
it will be from its negative terminal to its positive terminal. Whenever a battery is delivering energy to a circuit,
it will carry current in this direction. On the other hand,
when another source of emf is charging the battery in
question, it will have a current pushed through it from its
positive terminal to its negative terminal.
*Q28.2 The terminal potential difference is e – Ir where I is the
current in the battery in the direction from its negative
to its positive pole. So the answer to (i) is (d) and the
answer to (ii) is (b). The current might be zero or an
outside agent might push current backward through the
battery from positive to negative terminal.
*Q28.3
*Q28.4 Answers (b) and (d), as described by Kirchhoff’s junction rule.
*Q28.5 Answer (a).
Q28.6
The whole wire is very nearly at one uniform potential. There is essentially zero difference in
potential between the bird’s feet. Then negligible current goes through the bird. The resistance
through the bird’s body between its feet is much larger than the resistance through the wire
between the same two points.
*Q28.7 Answer (b). Each headlight’s terminals are connected to the positive and negative terminals of
the battery. Each headlight can operate if the other is burned out.
121
122
Q28.8
Chapter 28
Answer their question with a challenge. If the student is just looking at a diagram, provide the
materials to build the circuit. If you are looking at a circuit where the second bulb really is
fainter, get the student to unscrew them both and interchange them. But check that the student’s
understanding of potential has not been impaired: if you patch past the first bulb to short it out,
the second gets brighter.
*Q28.9 Answer (a). When the breaker trips to off, current does not go through the device.
*Q28.10 (i) For both batteries to be delivering electric energy, currents are in the direction g to a to b, h to
d to c, and so e to f. Points f, g, and h are all at zero potential. Points b, c, and e are at the same
higher voltage, d still higher, and a highest of all. The ranking is a > d > b = c = e > f = g = h.
(ii) The current in ef must be the sum of the other two currents. The ranking is e = f > g = a = b >
h = d = c.
*Q28.11 Closing the switch removes lamp C from the circuit, decreasing the resistance seen by the battery,
and so increasing the current in the battery. (i) Answer (a). (ii) Answer (d). (iii) Answer (a).
(iv) Answer (a). (v) Answer (d). (vi) Answer (a).
*Q28.12 Closing the switch lights lamp C. The action increases the battery current so it decreases the
terminal voltage of the battery. (i) Answer (b). (ii) Answer (a). (iii) Answer (a).
(iv) Answer (b). (v) Answer (a). (vi) Answer (a).
Q28.13 Two runs in series:
of one lift and two runs:
. Three runs in parallel:
. Junction
.
Gustav Robert Kirchhoff, Professor of Physics at Heidelberg and Berlin, was master of the
obvious. A junction rule: The number of skiers coming into any junction must be equal to the
number of skiers leaving. A loop rule: the total change in altitude must be zero for any skier
completing a closed path.
Q28.14 The bulb will light up for a while immediately after the switch is closed. As the capacitor charges,
the bulb gets progressively dimmer. When the capacitor is fully charged the current in the circuit
is zero and the bulb does not glow at all. If the value of RC is small, this whole process might
occupy a very short time interval.
Q28.15 The hospital maintenance worker is right. A hospital room is full of electrical grounds,
including the bed frame. If your grandmother touched the faulty knob and the bed frame at the
same time, she could receive quite a jolt, as there would be a potential difference of 120 V
across her. If the 120 V is DC, the shock could send her into ventricular fibrillation, and the
hospital staff could use the defibrillator you read about in Chapter 26. If the 120 V is AC,
which is most likely, the current could produce external and internal burns along the path of
conduction. Likely no one got a shock from the radio back at home because her bedroom
contained no electrical grounds—no conductors connected to zero volts. Just like the bird
in Question 28.6, granny could touch the “hot” knob without getting a shock so long as there was
no path to ground to supply a potential difference across her. A new appliance in the bedroom or a
flood could make the radio lethal. Repair it or discard it. Enjoy the news from Lake Wobegon on
the new plastic radio.
Q28.16 Both 120-V and 240-V lines can deliver injurious or lethal shocks, but there is a somewhat better
safety factor with the lower voltage. To say it a different way, the insulation on a 120-V line can
be thinner. On the other hand, a 240-V device carries less current to operate a device with the
same power, so the conductor itself can be thinner. Finally, as we will see in Chapter 33, the last
step-down transformer can also be somewhat smaller if it has to go down only to 240 volts from
the high voltage of the main power line.
Direct Current Circuits
123
SOLUTIONS TO PROBLEMS
Section 28.1
P28.1
(a)
(b)
Electromotive Force
P=
( ∆V )2
R
(11.6 V )2
becomes
20.0 W =
so
R = 6.73 Ω
∆V = IR
so
11.6 V = I ( 6.73 Ω )
and
I = 1.72 A
R
FIG. P28.1
ε = IR + Ir
15.0 V = 11.6 V + (1.72 A ) r
so
r = 1.97 Ω
P28.2
P28.3
The total resistance is R =
3.00 V
= 5.00 Ω.
0.600 A
(a)
Rlamp = R − rbatteries = 5.00 Ω − 0.408 Ω = 4.59 Ω
(b)
Pbatteries ( 0.408 Ω ) I 2
=
= 0.081 6 = 8.16%
Ptotal
( 5.00 Ω ) I 2
(a)
Here
ε = I (R + r) ,
so
I=
ε
R+r
=
FIG. P28.2
12.6 V
= 2.48 A.
( 5.00 Ω + 0.080 0 Ω )
Then, ∆V = IR = ( 2.48 A ) ( 5.00 Ω ) = 12.4 V .
(b)
*P28.4
Let I1 and I 2 be the currents flowing through the battery and the
headlights, respectively.
I1 = I 2 + 35.0 A , and ε − I1r − I 2 r = 0
Then,
ε = ( I 2 + 35.0 A ) ( 0.080 0 Ω ) + I 2 (5.00 Ω ) = 12.6 V
so
giving
I 2 = 1.93 A
Thus,
∆V2 = (1.93 A ) ( 5.00 Ω ) = 9.65 V
FIG. P28.3
(a)
At maximum power transfer, r = R. Equal powers are delivered to r and R. The efficiency
is 50.0% .
(b)
For maximum fractional energy transfer to R, we want zero energy absorbed by r, so we
want r = 0 .
(c)
High efficiency. The electric company’s economic interest is to minimize internal energy
production in its power lines, so that it can sell a large fraction of the energy output of its
generators to the customers.
(d)
High power transfer. Energy by electric transmission is so cheap compared to the sound
system that she does not spend extra money to buy an efficient amplifier.
124
Chapter 28
Section 28.2
P28.5
(a)
Resistors in Series and Parallel
Rp =
1
= 4.12 Ω
(1 7.00 Ω ) + (1 10.0 Ω )
Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω
(b)
∆V = IR
34.0 V = I (17.1 Ω )
I = 1.99 A
FIG. P28.5
for 4.00 Ω, 9.00 Ω resistors
Applying ∆V = IR,
(1.99 A ) ( 4.12 Ω ) = 8.18 V
8.18 V = I ( 7.00 Ω )
so
I = 1.17 A
for 7.00 Ω resistor
8.18 V = I (10.0 Ω )
so
*P28.6
I = 0.818 A
for 10.0 Ω resistor
(a)
The conductors in the cord have resistance. There is a potential
difference across each when current is flowing. The potential
difference applied to the light bulb is less than 120 V, so it will
carry less current than it is designed to, and will operate at
lower power than 75 W.
(b)
If the temperature of the bulb does not change much between
the design operating point and the actual operating point, we can
take the resistance of the filament as constant.
For the bulb in use as intended,
I=
and
R=
P
∆V
=
75.0 W
= 0.625 A
120 V
∆V
120 V
=
= 192 Ω
I
0.625 A
Now, presuming the bulb resistance is unchanged,
I=
120 V
= 0.620 A
193.6 Ω
Across the bulb is
∆V = IR = 192 Ω ( 0.620 A ) = 119 V
so its power is
P = I ∆V = 0.620 A (119 V ) = 73.8 W
FIG. P28.6
Direct Current Circuits
P28.7
125
If we turn the given diagram on its side, we find that it is the same as
figure (a). The 20.0 Ω and 5.00 Ω resistors are in series, so the first
reduction is shown in (b). In addition, since the 10.0 Ω, 5.00 Ω, and
25.0 Ω resistors are then in parallel, we can solve for their equivalent
resistance as:
1
= 2.94 Ω
Req =
1
1 ⎞
⎛ 1
+
+
⎜⎝ 10.0 Ω 5.00 Ω 25.0 Ω ⎟⎠
This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
∆V
Next, we work backwards through the diagrams applying I =
and
R
∆V = IR alternately to every resistor, real and equivalent. The 12.94 Ω
resistor is connected across 25.0 V, so the current through the battery in
every diagram is
I=
∆V 25.0 V
=
= 1.93 A
R 12.94 Ω
In figure (c), this 1.93 A goes through the 2.94 Ω equivalent resistor to
give a potential difference of:
∆V = IR = (1.93 A ) ( 2.94 Ω ) = 5.68 V
From figure (b), we see that this potential difference is the same across
∆Vab , the 10 Ω resistor, and the 5.00 Ω resistor.
Thus we have first found the answer to part (b), which is ∆Vab = 5.68 V .
(a)
I=
P28.8
FIG. P28.7
Since the current through the 20.0 Ω resistor is also the current
through the 25.0 Ω line ab,
∆Vab 5.68 V
=
= 0.227 A = 227 mA
Rab
25.0 Ω
We assume that the metal wand makes low-resistance contact with the person’s hand and that the
resistance through the person’s body is negligible compared to the resistance Rshoes of the shoe
soles. The equivalent resistance seen by the power supply is 1.00 MΩ + Rshoes . The current
50.0 V
. The voltmeter displays
through both resistors is
1.00 MΩ + Rshoes
∆V = I (1.00 MΩ ) =
(a)
We solve to obtain
50.0 V (1.00 MΩ ) = ∆V (1.00 MΩ ) + ∆V ( Rshoes )
Rshoes =
(b)
50.0 V (1.00 MΩ )
= ∆V
1.00 MΩ + Rshoes
1.00 MΩ ( 50.0 − ∆V )
∆V
With Rshoes → 0, the current through the person’s body is
50.0 V
= 50.0 µ A
1.00 MΩ
The current will never exceed 50 µ A.
126
P28.9
Chapter 28
(a)
Since all the current in the circuit must pass through the series 100 Ω
resistor, P = I 2 R
2
Pmax = RI max
FIG. P28.9
I max =
so
P
R
=
25.0 W
= 0.500 A
100 Ω
1
1 ⎞
Req = 100 Ω + ⎛
+
⎝ 100 100 ⎠
−1
Ω = 150 Ω
∆Vmax = Req I max = 75.0 V
(b)
P1 = I ∆V = ( 0.500 A ) ( 75.0 V ) = 37.5 W total power
P1 = 25.0 W
P2 = P3 = RI 2 (100 Ω ) ( 0.250 A )2 = 6.25 W
P28.10
Using 2.00-Ω, 3.00-Ω, and 4.00-Ω resistors,
there are 7 series, 4 parallel, and 6 mixed
combinations:
Series
2.00 Ω
3.00 Ω
4.00 Ω
5.00 Ω
6.00 Ω
7.00 Ω
9.00 Ω
Parallel
0.923 Ω
1.20 Ω
1.33 Ω
1.71 Ω
Mixed
1.56 Ω
2.00 Ω
2.22 Ω
3.71 Ω
4.33 Ω
5.20 Ω
FIG. P28.10
P28.11
When S is open, R1 , R2 , R3 are in series with the battery. Thus:
R1 + R2 + R3 =
6V
= 6 kΩ
10 −3 A
(1)
When S is closed in position 1, the parallel combination of the two R2’s is in series with R1 , R3 ,
and the battery. Thus:
R1 +
1
6V
= 5 kΩ
R2 + R3 =
2
1.2 × 10 −3 A
(2)
When S is closed in position 2, R1 and R2 are in series with the battery. R3 is shorted. Thus:
R1 + R2 =
6V
= 3 kΩ
2 × 10 −3 A
From (1) and (3): R3 = 3 kΩ.
Subtract (2) from (1): R2 = 2 kΩ.
From (3): R1 = 1 kΩ.
Answers: R1 = 1.00 kΩ, R2 = 2.00 kΩ, R3 = 3.00 kΩ .
(3)
Direct Current Circuits
P28.12
127
Denoting the two resistors as x and y,
1
1 1
= +
150 x y
x + y = 690, and
(690 − x ) + x
1
1
1
= +
=
x (690 − x )
150 x 690 − x
x 2 − 690 x + 103 500 = 0
690 ± ( 690 ) − 414 000
2
2
x=
x = 470 Ω
y = 220 Ω
*P28.13 The resistance between a and b decreases . Closing the switch opens a new path with resistance
of only 20 Ω. The original resistance is R +
1
1
90 + 10
1
The final resistance is R +
1
90
+
1
+
10
We require R + 50 Ω = 2(R + 18 Ω)
*P28.14 (a)
1
1
10
+
1
+
1
= R + 50 Ω.
10 + 90
= R + 18 Ω.
90
so R = 14.0 Ω
The resistors 2, 3, and 4 can be combined to a single 2R resistor. This is in series with
resistor 1, with resistance R, so the equivalent resistance of the whole circuit is 3R. In
1
series, potential difference is shared in proportion to the resistance, so resistor 1 gets
3
2
of the battery voltage and the 2-3-4 parallel combination gets of the battery voltage.
3
This is the potential difference across resistor 4, but resistors 2 and 3 must share this
1
2
voltage. In this branch goes to 2 and
to 3. The ranking by potential difference
3
3
is ∆V4 > ∆V3 > ∆V1 > ∆V2 .
(b)
Based on the reasoning above the potential differences
are ∆V1 =
ε , ∆V
3
2
=
2ε
4ε
2ε
, ∆V3 =
, ∆V4 =
.
9
9
3
(c)
All the current goes through resistor 1, so it gets the most. The current then splits at the
parallel combination. Resistor 4 gets more than half, because the resistance in that branch
is less than in the other branch. Resistors 2 and 3 have equal currents because they are in
series. The ranking by current is I1 > I 4 > I 2 = I 3 .
(d)
Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice that of
resistor 4, twice as much current goes through 4 as through 2 and 3. The currents through
I
2I
the resistors are I1 = I , I 2 = I 3 = , I 4 =
.
3
3
(e)
Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current in
the battery, which is the current through resistor 1, decreases. This decreases the potential
difference across resistor 1, increasing the potential difference across the parallel combination. With a larger potential difference the current through resistor 4 is increased. With
continued on next page
128
Chapter 28
more current through 4, and less in the circuit to start with, the current through resistors
2 and 3 must decrease. To summarize, I 4 increases and I1 , I 2 , and I 3 decrease .
(f)
If resistor 3 has an infinite resistance it blocks any current from passing through that
branch, and the circuit effectively is just resistor 1 and resistor 4 in series with the
battery. The circuit now has an equivalent resistance of 4R. The current in the
3
4
circuit drops to
of the original current because the resistance has increased by .
4
3
All this current passes through resistors 1 and 4, and none passes through 2 or 3.
3I
3I
, I 2 = I 3 = 0, I 4 =
.
4
4
Therefore I1 =
P28.15
1
1 ⎞
Rp = ⎛
+
⎝ 3.00 1.00 ⎠
−1
= 0.750 Ω
Rs = ( 2.00 + 0.750 + 4.00 ) Ω = 6.75 Ω
I battery =
∆V 18.0 V
=
= 2.67 A
Rs 6.75 Ω
P2 = ( 2.67 A )2 ( 2.00 Ω )
P2 = I 2 R:
P2 = 14.2 W in 2.00 Ω
P4 = ( 2.67 A )2 ( 4.00 A ) = 28.4 W in 4.00 Ω
∆V2 = ( 2.67 A ) ( 2.00 Ω ) = 5.33 V,
∆V4 = ( 2.67 A ) ( 4.00 Ω ) = 10.67 V
∆Vp = 18.0 V − ∆V2 − ∆V4 = 2.00 V ( = ∆V3 = ∆V1 )
P3 =
( ∆V3 )
P1 =
( ∆V1 ) = ( 2.00 V )2 =
Section 28.3
P28.16
R3
2
=
R1
( 2.00 V )2
3.00 Ω
1.00 Ω
FIG. P28.15
= 1.33 W in 3.000 Ω
4.00 W in 1.00 Ω
Kirchhoff’s Rules
+15.0 − ( 7.00 ) I1 − ( 2.00 ) ( 5.00 ) = 0
5.00 = 7.00 I1
so
I1 = 0.714 A
so
I 2 = 1.29 A
I 3 = I1 + I 2 = 2.00 A
0.714 + I 2 = 2.00
+ε − 2.00 (1.29 ) − 5.00 ( 2.00 ) = 0
ε = 12.6 V
FIG. P28.16
Direct Current Circuits
P28.17
We name currents I1 , I 2 , and I 3 as shown.
From Kirchhoff’s current rule, I 3 = I1 + I 2 .
Applying Kirchhoff’s voltage rule to the loop containing I 2 and I 3 ,
12.0 V − ( 4.00 ) I 3 − ( 6.00 ) I 2 − 4.00 V = 0
8.00 = ( 4.00 ) I 3 + ( 6.00 ) I 2
Applying Kirchhoff’s voltage rule to the loop containing I1 and I 2 ,
− ( 6.00 ) I 2 − 4.00 V + (8.00 ) I1 = 0
FIG. P28.17
(8.00 ) I1 = 4.00 + ( 6.00 ) I 2
Solving the above linear system, we proceed to the pair of simultaneous equations:
⎧8 = 4 I1 + 4 I 2 + 6 I 2
⎨
⎩8 I1 = 4 + 6 I 2
⎧8 = 4 I1 + 10 I 2
⎨
⎩ I 2 = 1.33 I1 − 0.667
or
and to the single equation 8 = 4 I1 + 13.3 I1 − 6.67
I1 =
and
14.7 V
= 0.846 A
17.3 Ω
I 3 = I1 + I 2
Then
give
I 2 = 1.33 ( 0.846 A ) − 0.667
I1 = 846 mA, I 2 = 462 mA, I 3 = 1.31 A
All currents are in the directions indicated by the arrows in the circuit diagram.
P28.18
The solution figure is shown to the right.
FIG. P28.18
*P28.19 We use the results of Problem 28.17.
(a)
(b)
(c)
By the 4.00-V battery:
∆U = ( ∆V ) I ∆t = ( 4.00 V ) ( −0.462 A )120 s = −222 J
By the 12.0-V battery:
(12.0 V ) (1.31 A )120 s = 1.88 kJ
By the 8.00-Ω resistor:
I 2 R∆t = ( 0.846 A ) (8.00 Ω )120 s = 687 J
By the 5.00-Ω resistor:
( 0.462 A )2 ( 5.00 Ω )120 s = 128 J
By the 1.00-Ω resistor:
( 0.462 A )2 (1.00 Ω )120 s = 25.6 J
By the 3.00-Ω resistor:
(1.31 A )2 ( 3.00 Ω )120 s = 616 J
By the 1.00-Ω resistor:
(1.31 A )2 (1.00 Ω )120 s = 205 J
2
−222 J + 1.88 kJ = 1.66 kJ from chemical to electrically transmitted. Like
a child counting his lunch money twice, we can count the same energy again,
687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ, as it is transformed from electrically
transmitted to internal. The net energy transformation is from chemical to internal .
129
130
Chapter 28
*P28.20 (a)
The first equation represents Kirchhoff’s loop
theorem. We choose to think of it as describing
a clockwise trip around the left-hand loop in a
circuit; see Figure (a). For the right-hand loop
see Figure (b). The junctions must be between
the 5.8 V and the 370 Ω and between the
370 Ω and the 150 Ω. Then we have
Figure (c). This is consistent with the third
equation,
I1
I2
220 Ω
Figure (a)
I3
I2
370 Ω
3.10 V
150 Ω
I1 + I 3 − I 2 = 0
Figure (b)
I 2 = I1 + I 3
I3
5.8 V
(b)
370 Ω
5.8 V
We substitute:
−220 I1 + 5.8 − 370 I1 − 370 I 3 = 0
I1
220 Ω
I2
+370 I1 + 370 I 3 + 150 I 3 − 3.1 = 0
150 Ω
370 Ω 3.10 V
Figure (c)
FIG. P28.20
Next
5.8 − 590 I1
370
520
370 I1 +
(5.8 − 590 I1 ) − 3.1 = 0
370
370 I1 + 8.15 − 829 I1 − 3.1 = 0
I3 =
I1 =
11.0 mA in the 220- Ω resistor and out of
5.05 V
=
the positive pole of the 5.8-V battery
459 Ω
I3 =
5.8 − 590 ( 0.011 0 )
= −1.87 mA
370
The current is 1.87 mA in the 150- Ω resistor and out of the
negative pole of the 3.1-V battery.
I 2 = 11.0 − 1.87 = 9.13 mA in the 370- Ω resistor
*P28.21 Let I6 represent the current in the ammeter and the top 6-Ω resistor. The bottom 6-Ω resistor has
the same potential difference across it, so it carries an equal current. For the top loop we have
6 V – 10 Ω I10 – 6 Ω I6 = 0
For the bottom loop, 4.5 – 5 I5 – 6 I6 = 0.
For the junctions on the left side, taken together, + I10 + I5 – I6 – I6 = 0.
We eliminate I10 = 0.6 – 0.6 I6
0.6 – 0.6 I6 + 0.9 – 1.2 I6 – 2 I6 = 0
and
I5 = 0.9 – 1.2 I6 by substitution:
I6 = 1.5/3.8 = 0.395 A
The loop theorem for the little loop containing the voltmeter gives
+ 6 V –∆V – 4.5 V = 0
∆V = 1.50 V
Direct Current Circuits
P28.22
Label the currents in the branches as shown in the first figure.
Reduce the circuit by combining the two parallel resistors as
shown in the second figure.
Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain:
( 2.71R ) I1 + (1.71R ) I 2 = 250
(1.71R ) I1 + ( 3.71R ) I 2 = 500
and
Figure (a)
With R = 1 000 Ω, simultaneous solution of these equations
yields:
I1 = 10.0 mA
and
I 2 = 130.0 mA
From Figure (b),
Vc − Va = ( I1 + I 2 ) (1.71R ) = 240 V
Thus, from Figure (a),
V − Va
240 V
I4 = c
=
= 60.0 mA
4R
4 000 Ω
Figure (b)
FIG. P28.22
Finally, applying Kirchhoff’s point rule at point a in Figure (a) gives:
I = I 4 − I1 = 60.0 mA − 10.0 mA = +50.0 mA
I = 50.0 mA from point a to point e
or
P28.23
Name the currents as shown in the figure to the right. Then w + x + z = y.
Loop equations are
−200 w − 40.0 + 80.0 x = 0
−80.0 x + 40.0 + 360 − 20.0 y = 0
+360 − 20.0 y − 70.0 z + 80.0 = 0
FIG. P28.23
Eliminate y by substitution.
⎧ x = 2.50 w + 0.500
⎪
⎨400 − 100 x − 20.0 w − 20.0 z = 0
⎪440 − 20.0 w − 20.0 x − 90.0 z = 0
⎩
Eliminate x.
⎧350 − 270 w − 20.0 z = 0
⎨
⎩430 − 70.0 w − 90.0 z = 0
Eliminate z = 17.5 − 13.5w to obtain
430 − 70.0 w − 1 575 + 1 215w = 0
w=
Now
70.0
= 1.00 A upward in 200 Ω
70.0
z = 4.00 A upward in 70.0 Ω
x = 3.00 A upward in 80.0 Ω
y = 8.00 A downward in 20.0 Ω
and for the 200 Ω,
∆V = IR = (1.00 A ) ( 200 Ω ) = 200 V
131
132
P28.24
Chapter 28
Using Kirchhoff’s rules,
12.0 − ( 0.010 0 ) I1 − ( 0.060 0 ) I 3 = 0
10.0 + (1.00 ) I 2 − ( 0.060 0 ) I 3 = 0
and
I1 = I 2 + I 3
12.0 − ( 0.010 0 ) I 2 − ( 0.070 0 ) I 3 = 0
FIG. P28.24
10.0 + (1.00 ) I 2 − ( 0.060 0 ) I 3 = 0
Solving simultaneously,
I 2 = 0.283 A downward in the dead battery
and
I 3 = 171 A downward in the starter
The currents are forward in the live battery and in the starter, relative to normal starting
operation. The current is backward in the dead battery, tending to charge it up.
P28.25
We name the currents I1 , I 2 , and I 3 as shown.
(a)
I1 = I 2 + I 3
Counterclockwise around the top loop,
12.0 V − ( 2.00 Ω ) I 3 − ( 4.00 Ω ) I1 = 0
Traversing the bottom loop,
8.00 V − ( 6.00 Ω ) I 2 + ( 2.00 Ω ) I 3 = 0
I1 = 3.00 −
(b)
4 1
1
I 3 , I 2 = + I 3 , and I 3 = 909 mA
3 3
2
FIG. P28.25
Va − ( 0.909 A ) ( 2.00 Ω ) = Vb
Vb − Va = −1.82 V
P28.26
∆Vab = (1.00 ) I1 + (1.00 ) ( I1 − I 2 )
∆Vab = (1.00 ) I1 + (1.000 ) I 2 + ( 5.00 ) ( I − I1 + I 2 )
∆Vab = ( 3.00 ) ( I − I1 ) + ( 5.00 ) ( I − I1 + I 2 )
Let I = 1.00 A, I1 = x , and I 2 = y.
FIG. P28.26
Then, the three equations become:
∆Vab = 2.00 x − y, or y = 2.00 x − ∆Vab
∆Vab = −4.00 x + 6.00 y + 5.00
and ∆Vab = 8.00 − 8.00 x + 5.00 y
Substituting the first into the last two gives:
7.00 ∆Vab = 8.00 x + 5.00
and
6.00 ∆Vab = 2.00 x + 8.00
Solving these simultaneously yields ∆Vab =
Then, Rab =
27
∆Vab
V
= 17
I
1.00 A
or
27
V.
17
Rab =
27
Ω
17
Direct Current Circuits
Section 28.4
P28.27
RC Circuits
(a)
RC = (1.00 × 10 6 Ω ) ( 5.00 × 10 −6 F ) = 5.00 s
(b)
Q = Cε = ( 5.00 × 10 −6 C ) ( 30.0 V ) = 150 µ C
(c)
I (t ) =
ε e− t RC = ⎛
⎡
⎤
−10.0
30.0 ⎞
exp ⎢
6
−6 ⎥
⎝ 1.00 × 10 6 ⎠
⎢⎣ (1.00 × 10 ) ( 5.00 × 10 ) ⎥⎦
R
FIG. P28.27
= 4.06 µ A
P28.28
The potential difference across the capacitor
∆V ( t ) = ∆Vmax (1 − e− t RC )
Therefore,
4.00 V = (10.0 V ) ⎡1 − e
⎢⎣
−( 3.00 ×105 Ω )
0.400 = 1.00 − e
Or
e
Taking the natural logarithm of both sides,
−
and
R=−
Using 1 Farad = 1 s Ω ,
P28.29
(a)
(
) ⎤
⎥⎦
−( 3.00 s ) ⎡⎣ R 10.0 ×10 −6 s Ω ⎤⎦
R
) = 0.600
(
− 3.00 ×105 Ω R
3.00 × 10 5 Ω
= ln ( 0.600 )
R
3.00 × 10 5 Ω
= +5.87 × 10 5 Ω = 587 kΩ
ln ( 0.600 )
I ( t ) = − I 0 e− t RC
I0 =
5.10 × 10 −6 C
Q
=
= 1.96 A
RC (1 300 Ω ) ( 2.00 × 10 −9 F )
⎤
⎡
−9.00 × 10 −6 s
I ( t ) = − (1.96 A ) exp ⎢
⎥ = −61.6 mA
−9
⎢⎣ (1 300 Ω ) ( 2.00 × 10 F ) ⎥⎦
P28.30
(b)
⎡
⎤
−8.00 × 10 −6 s
q ( t ) = Qe− t RC = ( 5.10 µ C ) exp ⎢
⎥ = 0.235 µ C
−9
⎢⎣ (1 300 Ω ) ( 2.00 × 10 F ) ⎦⎥
(c)
The magnitude of the maximum current is I 0 = 1.96 A .
We are to calculate
⬁
−2 t RC
∫ e dt = −
0
133
⬁
RC −2t RC ⎛ 2dt ⎞
RC −2t RC
e
−
=−
e
∫
⎝
⎠
2 0
2
RC
⬁
0
=−
RC − ⬁
RC
RC
⎡⎣ e − e0 ⎤⎦ = −
[ 0 − 1] = +
2
2
2
134
P28.31
Chapter 28
(a)
Call the potential at the left junction VL and at the right VR . After a
“long” time, the capacitor is fully charged.
VL = 8.00 V because of voltage divider:
10.0 V
= 2.00 A
5.00 Ω
VL = 10.0 V − ( 2.00 A ) (1.00 Ω ) = 8.00 V
IL =
Likewise,
2.00 Ω
⎞ (10.0 V ) = 2.00 V
VR = ⎛
⎝ 2.00 Ω + 8.00 Ω ⎠
or
IR =
FIG. P28.31(a)
10.0 V
= 1.00 A
10.0 Ω
VR = (10.0 V ) − (8.00 Ω ) (1.00 A ) = 2.00 V
(b)
Therefore,
∆V = VL − VR = 8.00 − 2.00 = 6.00 V
Redraw the circuit
R=
1
(1 9.00 Ω ) + (1 6.00 Ω )
= 3.60 Ω
RC = 3.60 × 10 −6 s
P28.32
1
10
and
e− t RC =
so
t = RC ln 10 = 8.29 µs
(a)
τ = RC = (1.50 × 10 5 Ω ) (10.0 × 10 −6 F ) = 1.50 s
(b)
τ = (1.00 × 10 5 Ω ) (10.0 × 10 −6 F ) = 1.00 s
(c)
The battery carries current
10.0 V
= 200 µ A
50.0 × 10 3 Ω
10.0 V ⎞ − t 1.00 s
I = I 0 e− t RC = ⎛
e
⎝ 100 × 10 3 Ω ⎠
The 100 kΩ carries current of magnitude
200 µ A + (100 µ A ) e− t 1.00 s
So the switch carries downward current
Section 28.5
P28.33
FIG. P28.31(b)
Electrical Meters
(
)
∆V = I g rg = I − I g Rp ,
or
Rp =
I g rg
(I − I )
g
=
I g ( 60.0 Ω )
(I − I )
g
Therefore, to have I = 0.100 A = 100 mA when I g = 0.500 mA:
Rp =
( 0.500 mA ) ( 60.0 Ω )
99.5 mA
= 0.302 Ω
FIG. P28.33
Direct Current Circuits
P28.34
135
)
(
Ammeter:
I g r = 0.500 A − I g ( 0.220 Ω )
or
I g ( r + 0.220 Ω ) = 0.110 V
(1)
Voltmeter:
2.00 V = I g ( r + 2 500 Ω )
(2)
Solve (1) and (2) simultaneously to find:
I g = 0.756 mA
P28.35
and
r = 145 Ω
FIG. P28.34
Series Resistor → Voltmeter
∆V = IR:
25.0 = 1.50 × 10 −3 ( Rs + 75.0 )
Solving,
Rs = 16.6 kΩ
FIG. P28.35
*P28.36 (a)
In Figure (a), the
emf sees an
equivalent
resistance
of 200.00 Ω.
I=
20.000 Ω
20.000 Ω
6.0000 V
6.000 0 V
200.00 Ω
= 0.030 000 A
20.000 Ω
A
V
V
A
180.00 Ω
180.00 Ω
180.00 Ω
(a)
(b)
(c)
FIG. P28.36
The terminal potential difference is
∆V = IR = ( 0.030 000 A ) (180.00 Ω ) = 5.400 0 V
(b)
⎛
1
1
⎞
Req = ⎜
+
⎝ 180.00 Ω 20 000 Ω ⎟⎠
(c)
In Figure (b),
−1
= 178.39 Ω
The equivalent resistance across
the emf is
178.39 Ω + 0.500 00 Ω + 20.000 Ω = 198.89 Ω
The ammeter reads
I=
and the voltmeter reads
∆V = IR = ( 0.030 167 A ) (178.39 Ω ) = 5.381 6 V
In Figure (c),
⎛
1
1
⎞
⎜⎝ 180.50 Ω + 20 000 Ω ⎟⎠
Therefore, the emf sends current
through
ε = 6.000 0 V =
R
198.89 Ω
0.030 167 A
−1
= 178.89 Ω
Rtot = 178.89 Ω + 20.000 Ω = 198.89 Ω
6.000 0 V
= 0.030 168 A
198.89 Ω
The current through the battery is
but not all of this goes through the
ammeter.
I=
The voltmeter reads
∆V = IR = ( 0.030 168 A ) (178.89 Ω ) = 5.396 6 V
The ammeter measures current
I=
∆V 5.396 6 V
=
= 0.029 898 A .
R
180.50 Ω
continued on next page
136
(d)
Chapter 28
Both circuits are good enough for some measurements. The connection in Figure (c) gives data
leading to value of resistance that is too high by only about 0.3%. Its value is more accurate than
the value, 0.9% too low, implied by the data from the circuit in part (b).
Section 28.6
P28.37
(a)
Household Wiring and Electrical Safety
P = I ∆V :
P 1 500 W
=
= 12.5 A
∆V
120 V
750 W
I=
= 6.25 A
120 V
1 000 W
I=
= 8.33 A
120 V
I=
So for the Heater,
For the Toaster,
And for the Grill,
P28.38
(b)
12.5 + 6.25 + 8.33 = 27.1 A
The current draw is greater than 25.0 amps, so a circuit with this circuit breaker would not
be sufficient.
(a)
Suppose that the insulation between either of your fingers and the conductor adjacent is a
chunk of rubber with contact area 4 mm 2 and thickness 1 mm. Its resistance is
13
−3
ρ (10 Ω ⋅ m ) (10 m )
≈ 2 × 1015 Ω
≈
4 × 10 −6 m 2
A
The current will be driven by 120 V through total resistance (series)
R=
2 × 1015 Ω + 10 4 Ω + 2 × 1015 Ω ≈ 5 × 1015 Ω
∆V
120 V
~
~ 10 −14 A .
R 5 × 1015 Ω
Vh
, where
The resistors form a voltage divider, with the center of your hand at potential
2
Vh is the potential of the “hot” wire. The potential difference between your finger and
It is: I =
(b)
thumb is ∆V = IR ~ (10 −14 A ) (10 4 Ω ) ~ 10 −10 V. So the points where the rubber meets your
fingers are at potentials of
~
Vh
+ 10 −10 V
2
and
~
Vh
− 10 −10 V
2
Additional Problems
*P28.39 Several seconds is many time constants, so the capacitor is fully charged and (d) the current in its
branch is zero .
Center loop:
+8 V + 3 Ω I2 – 5 Ω I1 = 0
Right loop:
+4 V – 3 Ω I2 – 5 Ω I3 = 0
Top junction:
+ I1 + I2 – I3 = 0
Now we will eliminate I1 = 1.6 + 0.6I2
and I3 = 0.8 – 0.6I2
by substitution: 1.6 + 0.6I2 + I2 – 0.8 + 0.6I2 = 0 Then
I2 = –0.8/2.2 = –0.3636
So (b) the current in 3 Ω is 0.364 A down .
Now (a) I3 = 0.8 – 0.6(–0.364) = 1.02 A down in 4 V and in 5 Ω .
(c) I1 = 1.6 + 0.6(–0.364) = 1.38 A up in the 8 V battery
(e) For the left loop +3 V – Q/6 m F + 8 V = 0
so
Q = 6 mF 11 V = 66.0 mC
Direct Current Circuits
15 V
*P28.40 The current in the battery is
10 Ω +
= 1.15 A.
1
1
5Ω
+ 1
137
8Ω
The voltage across 5 Ω is 15 V – 10 Ω 1.15 A = 3.53 V.
P28.41
(a)
The current in it is 3.53 V/5 Ω = 0.706 A.
(b)
P = 3.53 V 0.706 A = 2.49 W
(c)
Only the circuit in Figure P28.40c requires the use of Kirchhoff’s rules for solution. In the
other circuits the 5-Ω and 8-Ω resistors are still in parallel with each other.
(d)
The power is lowest in Figure P28.40c. The circuits in Figures P28.40b and P28.40d have
in effect 30-V batteries driving the current.
The set of four batteries boosts the electric potential of each bit of charge that goes through them
by 4 × 1.50 V = 6.00 V. The chemical energy they store is
∆U = q∆V = ( 240 C ) ( 6.00 J C ) = 1 440 J
∆V 6.00 V
=
= 0.030 0 A
R
200 Ω
The radio draws current
I=
So, its power is
P = ( ∆V ) I = ( 6.00 V ) ( 0.030 0 A ) = 0.180 W = 0.180 J s
Then for the time the energy lasts,
E
we have P =
:
∆t
∆t =
We could also compute this from I =
*P28.42 I =
ε
R+r
Let x ≡
, so
P = I 2R =
ε 2R
(R + r)
ε 2 , then ( R + r )2 = xR
P
or
2
Q
:
∆t
∆t =
E
P
=
1 440 J
= 8.00 × 10 3 s
0.180 J s
Q
240 C
=
= 8.00 × 10 3 s = 2.22 h
I 0.030 0 A
⎛ ε2 ⎞
R
⎝ P ⎟⎠
( R + r )2 = ⎜
R 2 + ( 2r − x ) R − r 2 = 0
or
With r = 1.20 Ω, this becomes
R 2 + ( 2.40 − x ) R − 1.44 = 0
which has solutions of
R=
− ( 2.40 − x ) ± ( 2.40 − x ) − 5.76
2
2
(a)
With
ε = 9.20 V
P = 12.8 W, x = 6.61:
and
+4.21 ± ( 4.21) − 5.76
= 3.84 Ω or 0.375 Ω . Either external
2
resistance extracts the same power from the battery.
2
R=
(b)
For
ε = 9.20 V
P = 21.2 W, x ≡
and
ε 2 = 3.99
P
+1.59 ± (1.59 ) − 5.76 1.59 ± −3.22
=
2
2
2
R=
The equation for the load resistance yields a complex number, so there is no resistance
that will extract 21.2 W from this battery. The maximum power output occurs when
R = r = 1.20 Ω, and that maximum is
Pmax =
ε 2 = 17.6 W.
4r
138
P28.43
Chapter 28
Using Kirchhoff’s loop rule for the closed loop, +12.0 − 2.00 I − 4.00 I = 0, so I = 2.00 A
Vb − Va = +4.00 V − ( 2.00 A ) ( 4.00 Ω ) − ( 0 ) (10.0 Ω ) = −4.00 V
Thus, ∆Vab = 4.00 V and point a is at the higher potential .
P28.44
ε
Req = 3R
(b)
Req =
(c)
Nine times more power is converted in the parallel
I=
R
1
=
(1 R ) + (1 R ) + (1 R ) 3
*P28.45 The charging current is given by
I=
3R
3ε
R
ε2
Pseries = ε I =
(a)
3R
Pparallel = ε I =
3ε 2
R
connection.
14.7 V – 13.2 V – I 0.85 Ω = 0
I = 1.76 A
The energy delivered by the 14.7 V supply is ∆VIt = 14.7 V 1.76 A 1.80 h (3 600 s/h) = 168 000 J
The energy stored in the battery is
13.2 V 1.76 A 1.8 (3 600 s) = 151 000 J
The same energy is released by the emf of the battery:
13.2 V (I ) 7.3 (3 600 s) = 151 000 J
so the discharge current is I = 0.435 A
The load resistor is given by
13.2 V – (0.435 A) R – (0.435 A) 0.85 Ω = 0
R = (12.8 V)/0.435 A = 29.5 Ω
The energy delivered to the load is ∆VIt = I 2 R t = (0.435 A)2 29.5 Ω (7.3) 3 600 s = 147 000 J
The efficiency is 147 000 J/168 000 J = 0.873
P28.46
(a)
ε − I ( ∑ R ) − (ε1 + ε 2 ) = 0
40.0 V − ( 4.00 A )[( 2.00 + 0.300 + 0.300 + R ) Ω ] − ( 6.00 + 6.00 ) V = 0; so R = 4.40 Ω
(b)
(c)
Inside the supply,
P = I 2 R = ( 4.00 A )2 ( 2.00 Ω ) = 32.0 W
Inside both batteries together,
P = I 2 R = ( 4.00 A )2 ( 0.600 Ω ) = 9.60 W
For the limiting resistor,
P = ( 4.00 A )2 ( 4.40 Ω ) = 70.4 W
P = I (ε1 + ε 2 ) = ( 4.00 A )[( 6.00 + 6.00 ) V ] = 48.0 W
Direct Current Circuits
P28.47
Let the two resistances be x and y.
Ps
Rs = x + y =
=
and
Rp =
so
x ( 9.00 Ω − x )
= 2.00 Ω
x + ( 9.00 Ω − x )
I2
( 5.00 A )2
= 9.00 Ω
y = 9.00 Ω − x
x
P p 50.0 W
xy
= 2 =
= 2.00 Ω
x+y I
( 5.00 A )2
y
x 2 − 9.00 x + 18.0 = 0
FIG. P28.47
Factoring the second equation,
( x − 6.00 ) ( x − 3.00 ) = 0
so
x = 6.00 Ω
or
x = 3.00 Ω
y = 3.00 Ω
or
y = 6.00 Ω
Then,
y
x
225 W
Then,
y = 9.00 Ω − x gives
139
There is only one physical answer: The two resistances are 6.00 Ω and 3.00 Ω .
P28.48
Then, Rs = x + y =
Ps
From the first equation, y =
becomes
Rp =
and
I2
x (P s I 2 − x )
x + (P s I − x )
2
=
Ps
I2
I
2
Using the quadratic formula, x =
Then, y =
Ps
I2
− x gives y =
The two resistances are
P28.49
(a)
∆V1 = ∆V2
I = I1 + I 2 = I1 +
I1 =
(b)
IR2
R1 + R2
Pp
xy
=
.
x + y I2
x2 −
or
2I 2
Ps ∓ Ps2 − 4PsP p
2I 2
FIG. P28.48
.
.
Ps + Ps2 − 4PsP p
and
2I 2
Ps − Ps2 − 4PsP p
2I 2
.
I1 R1 = I 2 R2
I1 R1
R + R1
= I1 2
R2
R2
I2 =
dP
= 2 I1 R1 + 2 ( I − I1 ) ( −1) R2 = 0
dI1
R1
I1
R2
I2
I
I1 R1
IR1
=
= I2
R2
R1 + R2
power we want to find I1 such that
IR2
R1 + R2
y
PsP p
⎛ Ps ⎞
x + 4 = 0.
⎝ I2 ⎠
I
Ps ± Ps2 − 4PsP p
The power delivered to the pair is
I1 =
x
− x , and the second
Pp
y
x
Let the two resistances be x and y.
FIG. P28.49(a)
P = I12 R1 + I 22 R2 = I12 R1 + ( I − I1 ) R2 . For minimum
dP
2
dI1
= 0.
I1 R1 − IR2 + I1 R2 = 0
This is the same condition as that found in part (a).
140
Chapter 28
*P28.50 (a) When the capacitor is fully charged, no current exists in its branch. The current in the left
resistors is 5 V/83 Ω = 0.060 2 A. The current in the right resistors is 5 V/(2 Ω + R).
Relative to the negative side of the battery, the left capacitor plate is at voltage
80 Ω (0.060 2 A) = 4.82 V. The right plate is at R (5 V)/(2 Ω + R).
The voltage across the capacitor is 4.82 V – R (5 V)/(2 Ω + R).
The charge on the capacitor is
Q = 3 mF [4.82 V – R (5 V)/(2 Ω + R)] = (28.9 Ω – 0.542 R) mC/(2 Ω + R)
(b) With R = 10 Ω,
(c)
Q = (28.9 – 5.42) mC/(2 + 10) = 1.96 mC
Yes. Q = 0 when 28.9 Ω – 0.542 R = 0
R = 53.3 Ω
(d) The maximum charge occurs for R = 0. It is 28.9/2 = 14.5 m C .
(e)
Yes. Taking R = ∞ corresponds to disconnecting a wire to remove the branch containing R.
In this case |Q|= 0.542 R/R = 0.542 m C .
P28.51
Let Rm = measured value, R = actual value,
I R = current through the resistor R
I = current measured by the ammeter
(a)
When using circuit (a), I R R = ∆V = 20 000 ( I − I R ) or
⎡I
⎤
R = 20 000 ⎢ − 1⎥
⎣ IR
⎦
∆V
But since I =
and
Rm
IR =
∆V
, we have
R
( R − Rm )
and
R = 20 000
When R > Rm , we require
( R − Rm ) ≤ 0.050 0
Therefore, Rm ≥ R (1 − 0.050 0 ) and from (1) we find
(b)
FIG. P28.51
I
R
=
I R Rm
When using circuit (b),
But since I R =
∆V
,
Rm
When Rm > R, we require
From (2) we find
(1)
Rm
R
R ≤ 1 050 Ω
I R R = ∆V − I R ( 0.5 Ω )
Rm = ( 0.500 + R )
( Rm − R ) ≤ 0.050 0
R
R ≥ 10.0 Ω
(2)
Direct Current Circuits
P28.52
The battery supplies energy at a changing rate
dE
⎛ ε −1 RC ⎞
= P = εI = ε
e
⎝R
⎠
dt
Then the total energy put out by the battery is
∫ dE =
⬁
∫
t =0
141
ε 2 exp ⎛ −
t ⎞
dt
⎝ RC ⎠
R
ε 2 ( − RC ) ⬁ exp ⎛ − t ⎞ ⎛ − dt ⎞ = − ε 2C exp ⎛ − t ⎞ ⬁ = −ε 2C [ 0 − 1] = ε 2C
∫0 ⎝ RC ⎠ ⎝ RC ⎠
⎝ RC ⎠ 0
R
dE
ε 2 exp ⎛ − 2t ⎞
= P = ∆V I = I 2 R = R
The power delivered to the resistor is
∫ dE =
R
dt
∞
∫ dE = ∫
So the total internal energy appearing in the resistor is
0
∫ dE =
ε
is conserved
the battery.
(a)
R
2t ⎞
dt
⎝ RC ⎠
⬁
2
1
1
2
C ( ∆V ) = Cε 2 . Thus, energy of the circuit
2
2
ε 2C = 1 ε 2C + 1 ε 2C and resistor and capacitor share equally in the energy from
2
2
q = C ∆V (1 − e− t RC )
q = (1.00 × 10 −6 F ) (10.0 V ) ⎡1 − e
⎢⎣
(b)
ε 2 exp ⎛ −
2
2
⎛ − RC ⎞ exp ⎛ − 2t ⎞ ⎛ − 2dt ⎞ = − ε C exp ⎛ − 2t ⎞ = − ε C 0 − 1 = ε C
[
]
⎝ RC ⎠ ⎝ RC ⎠
⎝ RC ⎠ 0
R ⎝ 2 ⎠ ∫0
2
2
2
⬁
2
The energy finally stored in the capacitor is U =
P28.53
⎝ RC ⎠
R2
I=
(
)(
) ⎤
= 9.93 µ C
⎥⎦
−10.0 ⎡⎣ 2.00 ×106 1.00 ×10 −6 ⎤⎦
dq ⎛ ∆V ⎞ − t RC
=
e
dt ⎝ R ⎠
10.0 V ⎞ −5.00
I =⎛
e
= 3.37 × 10 −8 A = 33.7 nA
⎝ 2.00 × 10 6 Ω ⎠
(c)
dU d ⎛ 1 q 2 ⎞ ⎛ q ⎞ dq ⎛ q ⎞
=
I
=
=
dt dt ⎜⎝ 2 C ⎟⎠ ⎝ C ⎠ dt ⎝ C ⎠
dU ⎛ 9.93 × 10 −6 C ⎞
=
(3.37 × 10 −8 A ) = 3.34 × 10 −7 W = 334 nW
dt ⎜⎝ 1.00 × 10 −6 C V ⎟⎠
(d)
Pbattery = I ε = ( 3.37 × 10 −8 A ) (10.0 V ) = 3.37 × 10 −7 W = 337 nW
The battery power could also be computed as the sum of the instantaneous powers
delivered to the resistor and to the capacitor:
I 2R + dU/dt = (33.7 × 10−9 A)2 2 × 106 Ω + 334 nW = 337 nW
142
Chapter 28
*P28.54 (a)
0.9 Ω = r
We find the resistance intrinsic to the vacuum
cleaner:
P = I ∆V =
R=
( ∆V )2
P
( ∆V )2
120 V = e
R
=
(120 V )2
535 W
= 26.9 Ω
0.9 Ω = r
with the inexpensive cord, the equivalent
resistance is 0.9 Ω + 26.9 Ω + 0.9 Ω = 28.7 Ω
so the current throughout the circuit is
I=
ε
RTot
=
26.9 Ω = R
FIG. P28.54
120 V
= 4.18 A
28.7 Ω
and the cleaner power is
Pcleaner = I ( ∆V )cleaner = I 2 R = ( 4.18 A )2 ( 26.9 Ω ) = 470 W
In symbols, RTot = R + 2r , I =
(b)
⎛ ε 2R ⎞
R + 2r = ⎜
⎝ Pcleaner ⎟⎠
r=
and
R + 2r
26.9 Ω ⎞
= 120 V ⎛
⎝ 525 W ⎠
Pcleaner = I 2 R =
ε 2R
( R + 2 r )2
12
= 27.2 Ω
ρ ρ44
27.2 Ω − 26.9 Ω
= 0.128 Ω =
=
2
A π d2
4 ρ ⎞
d=⎛
⎝ πr ⎠
(c)
12
ε
12
⎛ 4 (1.7 × 10 −8 Ω ⋅ m ) (15 m ) ⎞
=⎜
⎟
π ( 0.128 Ω )
⎝
⎠
12
= 1.60 mm or more
Unless the extension cord is a superconductor, it is impossible to attain cleaner power
535 W. To move from 525 W to 532 W will require a lot more copper, as we show here:
r=
ε⎛
R ⎞
⎜
2 ⎝ Pcleaner ⎟⎠
4 ρ ⎞
d=⎛
⎝ πr ⎠
12
12
−
R 120 V ⎛ 26.9 Ω ⎞
=
2
2 ⎝ 532 W ⎠
12
−
⎛ 4 (1.7 × 10 −8 Ω ⋅ m ) (15 m ) ⎞
=⎜
⎟
π ( 0.037 9 Ω )
⎝
⎠
26.9 Ω
= 0.037 9 Ω
2
12
= 2.93 mm or more
Direct Current Circuits
P28.55
(a)
First determine the resistance of each light bulb:
R=
( ∆V )2
P
=
(120 V )2
60.0 W
P=
( ∆V )2
R
= 240 Ω
We obtain the equivalent resistance Req of the network of light
bulbs by identifying series and parallel equivalent resistances:
1
Req = R1 +
= 240 Ω + 120 Ω = 360 Ω
(1 R2 ) + (1 R3 )
The total power dissipated in the 360 Ω is
(b)
The current through the network is given by
The potential difference across R1 is
P=
( ∆V )2
Req
=
FIG. P28.55
(120 V )2
P = I 2 Req : I =
360 Ω
P
Req
=
= 40.0 W
40.0 W 1
= A
360 Ω
3
1
∆V1 = IR1 = ⎛ A⎞ ( 240 Ω ) = 80.0 V
⎝3 ⎠
The potential difference ∆V23 across the parallel combination of R2 and R3 is
⎛
⎞
1
1
∆V23 = IR23 = ⎛ A⎞ ⎜
= 40.0 V
⎝ 3 ⎠ ⎝ (1 240 Ω ) + (1 240 Ω ) ⎟⎠
P28.56
(a)
With the switch closed, current exists in a simple series
circuit as shown. The capacitors carry no current.
For R2 we have
P = I 2 R2
I=
P
R2
=
2.40 V ⋅ A
= 18.5 mA
7 000 V A
The potential difference across R1 and C1 is
∆V = IR1 = (1.85 × 10 −2 A ) ( 4 000 V A ) = 74.1 V
FIG. P28.56(a)
The charge on C1
Q = C1 ∆V = ( 3.00 × 10 −6 C V ) ( 74.1 V ) = 222 µ C
The potential difference across R2 and C2 is
∆V = IR2 = (1.85 × 10 −2 A ) ( 7 000 Ω ) = 130 V
The charge on C2
Q = C2 ∆V = ( 6.00 × 10 −6 C V ) (130 V ) = 778 µ C
The battery emf is
IReq = I ( R1 + R2 ) = 1.85 × 10 −2 A ( 4 000 + 7 000 ) V A = 204 V
(b)
In equilibrium after the switch has been opened, no current
exists. The potential difference across each resistor is zero.
The full 204 V appears across both capacitors. The new
charge on C2 is
Q = C2 ∆V = ( 6.00 × 10 −6 C V ) ( 204 V ) = 1 222 µ C
for a change of 1 222 µ C − 778 µ C = 444 µ C .
FIG. P28.56(b)
143
144
Chapter 28
*P28.57 (a) The emf of the battery is 9.30 V . Its internal resistance is given by
9.30 V – 3.70 A r = 0
r = 2.51 Ω
(b) Total emf = 20(9.30 V) = 186 V . The maximum current is given by
20(9.30 V) – 20(2.51 Ω) I – 0 I = 0
I = 3.70 A
(c)
For the circuit 20(9.30 V) – 20(2.51 Ω) I – 120 Ω I = 0
(d)
P = I R = (1.09 A) 120 Ω = 143 W . This is a potentially deadly situation.
(e)
The potential difference across his body is 120 Ω (0.005 00 A) = 0.600 V.
2
I = 186 V/170 Ω = 1.09 A
2
This must be the terminal potential difference of the bank of batteries:
186 V – Itot 20(2.51 Ω) = 0.6 V
Itot = 185.4 V/50.3 Ω = 3.688 A
For the copper wire we then have
0.6 V = (3.688 A – 0.005 A) R R = 0.163 Ω
(f)
For the experimenter’s body, P = I∆V = 0.005 A 0.6 V = 3.00 mW .
(g)
For the wire P = I∆V = 3.683 A 0.6 V = 2.21 W .
(h)
The power output of the emf depends on the resistance connected to it. A question about
“the rest of the power” is not meaningful when it compares circuits with different currents.
The net emf produces more current in the circuit where the copper wire is used. The net
emf delivers more power when the copper wire is used, 687 W rather than 203 W without
the wire. Nearly all of this power results in extra internal energy in the internal resistance
of the batteries, which rapidly rise to a high temperature. The circuit with the copper wire is
unsafe because the batteries overheat. The circuit without the copper wire is unsafe because
it delivers an electric shock to the experimenter.
*P28.58 The battery current is
(150 + 45 + 14 + 4 ) mA = 213 mA
(a)
The resistor with highest resistance is that
carrying 4 mA. Doubling its resistance
FIG. P28.58
will reduce the current it carries to 2 mA.
Then the total current is
211
= 0.991 .
(150 + 45 + 14 + 2 ) mA = 211 mA, nearly the same as before. The ratio is
213
(b)
The resistor with least resistance carries 150 mA. Doubling its resistance changes this
current to 75 mA and changes the total to ( 75 + 45 + 14 + 4 ) mA = 138 mA. The ratio
138
is
= 0.648 , representing a much larger reduction (35.2% instead of 0.9%).
213
(c)
This problem is precisely analogous. As a battery maintained a potential difference in
parts (a) and (b), a furnace maintains a temperature difference here. Energy flow by heat is
analogous to current and takes place through thermal resistances in parallel. Each resistance
can have its “R-value” increased by adding insulation. Doubling the thermal resistance of
the attic door will produce only a negligible (0.9%) saving in fuel. The ceiling originally
has the smallest thermal resistance. Doubling the thermal resistance of the ceiling will
produce a much larger saving.
Direct Current Circuits
P28.59
.
From the hint, the equivalent resistance of
That is,
RT +
1
= Req
1 RL + 1 Req
RT +
RL Req
RL + Req
= Req
RT RL + RT Req + RL Req = RL Req + Req2q
Req2 − RT Req − RT RL = 0
Req =
RT ± RT2 − 4 (1) ( − RT RL )
2 (1)
Only the + sign is physical:
Req =
P28.60
1
2
(
4 RT RL + RT2 + RT
For example, if
RT = 1 Ω
And
RL = 20 Ω, Req = 5 Ω
(a)
First let us flatten the circuit on a 2-D
plane as shown; then reorganize it to a
format easier to read. Notice that the two
resistors shown in the top horizontal
branch carry the same current as the
resistors in the horizontal branch
second from the top. The center
junctions in these two branches are at
the same potential. The vertical resistor
between these two junctions has no
potential difference across it and
carries no current. This middle resistor
can be removed without affecting the
circuit. The remaining resistors over
the three parallel branches have
equivalent resistance
1
1
1
Req = ⎛ +
+ ⎞
⎝ 20 20 10 ⎠
(b)
)
−1
= 5.00 Ω
So the current through the battery is
∆V 12.0 V
=
= 2.40 A
Req 5.00 Ω
FIG. P28.60(a)
145
146
P28.61
Chapter 28
(a)
After steady-state conditions have been reached, there is no DC current through the capacitor.
Thus, for R3 :
I R3 = 0 ( steady-state )
For the other two resistors, the steady-state current is simply determined by the 9.00-V emf
across the 12-kΩ and 15-kΩ resistors in series:
For R1 and R2 :
(b)
I ( R1 + R2 ) =
ε
R1 + R2
=
9.00 V
(12.0 kΩ + 15.0 kΩ )
= 333 µ A ( steady-state )
After the transient currents have ceased, the potential
difference across C is the same as the potential
difference across R2 ( = IR2 ) because there is no voltage
drop across R3 . Therefore, the charge Q on C is
Q = C ( ∆V )R2 = C ( IR2 ) = (10.0 µ F ) ( 333 µ A ) (15.0 kΩ )
= 50.0 µ C
FIG. P28.61(b)
(c)
When the switch is opened, the branch containing R1
is no longer part of the circuit. The capacitor discharges
through ( R2 + R3 ) with a time constant of
( R2 + R3 ) C = (15.0 kΩ + 3.00 kΩ ) (10.0 µF ) = 0.180 s.
The initial current I i in this discharge circuit is
determined by the initial potential difference across the
capacitor applied to ( R2 + R3 ) in series:
(333 µ A ) (15.0 kΩ ) = 278 µ A
IR2
Ii =
=
=
( R2 + R3 ) ( R2 + R3 ) (15.0 kΩ + 3.00 kΩ )
( ∆V )C
FIG. P28.61(c)
Thus, when the switch is opened, the current through R2 changes instantaneously from
333 µ A (downward) to 278 µ A (downward) as shown in the graph. Thereafter, it decays
according to
I R2 = I i e− t ( R2 + R3 )C =
(d)
( 278 µ A ) e− t (0.180 s) ( for t > 0 )
The charge q on the capacitor decays from Qi to
q = Qi e− t ( R2 + R3 )C
Qi
= Qi e( − t 0.180 s)
5
5 = et 0.180 s
ln 5 =
t
180 ms
t = ( 0.180 s ) ( ln 5) = 290 ms
Qi
according to
5
Direct Current Circuits
P28.62
147
∆V = ε e− t RC
so
ln
⎛ ε ⎞ ⎛ 1 ⎞
=
t
⎝ ∆V ⎠ ⎝ RC ⎠
⎛ ε ⎞
versus t
⎝ ∆V ⎠
should be a straight line with
1
slope equal to
.
RC
A plot of ln
Using the given data
values:
(a)
FIG. P28.62
A least-square fit to this data yields the graph above.
∑ x = 282,
∑ x y = 244,
∑x
∑y
i
Slope =
N
4
= 4.03,
i
i i
= 1.86 × 10 ,
2
i
N =8
(∑ x y ) − (∑ x )(∑ y ) = 0.011 8
N (∑ x ) − (∑ x )
i i
i
Intercept =
i
2
2
i
i
(∑ x )(∑ y ) − (∑ x )(∑ x y ) = 0.088 2
N (∑ x ) − (∑ x )
2
i
i
i
i i
2
2
i
i
The equation of the best fit line is:
(b)
P28.63
(a)
ln
t (s)
0
4.87
11.1
19.4
30.8
46.6
67.3
1022.2
⎛ ε ⎞
= ( 0.011 8 ) t + 0.088 2
⎝ ∆V ⎠
Thus, the time constant is
τ = RC =
and the capacitance is
C=
1
1
=
= 84.7 s
slope 0.011 8
τ
84.7 s
=
= 8.47 µ F
R 10.0 × 10 6 Ω
R1
For the first measurement, the equivalent circuit is as
shown in Figure 1.
Rab = R1 = Ry + Ry = 2 Ry
so
Ry =
1
R1
2
a
1
Rac = R2 = Ry + Rx
Thus,
2
Substitute (1) into (2) to obtain:
b
c
Ry
Ry
Rx
(1)
Figure 1
For the second measurement, the equivalent circuit is
shown in Figure 2.
R2
a
c
(2)
1
1 1
R2 = ⎛ R1 ⎞ + Rx , or Rx = R2 − R1
4
2⎝2 ⎠
(b)
ln (ε ∆V )
0
0.109
0..228
0.355
0.509
0.695
0.919
1.219
∆V (V )
6.19
5.55
4.93
4.34
3.72
3.09
2.47
1.83
If R1 = 13.0 Ω and R2 = 6.00 Ω, then Rx = 2.75 Ω .
Ry
Ry
Figure 2
FIG. P28.63
The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω
Rx
148
P28.64
Chapter 28
Start at the point when the voltage has just reached
2
∆V and the switch has just closed. The voltage
3
2
is ∆V and is decaying towards 0 V with a
3
time constant R2 C
+
R1
∆V
Voltage
controlled
switch
R2
V ∆Vc
C
2
∆VC ( t ) = ⎡⎢ ∆V ⎤⎥ e− t R2C
⎣3
⎦
We want to know when ∆VC ( t ) will reach
Therefore,
1
2
∆V = ⎡⎢ ∆V ⎤⎥ e− t R2C
3
⎣3
⎦
or
e− t R2C =
or
t1 = R2 C ln 2
1
∆V.
3
1
2
FIG. P28.64
1
After the switch opens, the voltage is ∆V , increasing toward ∆V with time constant
3
( R1 + R2 ) C:
2
∆VC ( t ) = ∆V − ⎡⎢ ∆V ⎤⎥ e− t ( R1 + R2 )C
⎣3
⎦
2
∆VC ( t ) = ∆V
3
2
2
∆V = ∆V − ∆Ve− t ( R1 + R2 )C
3
3
When
t 2 = ( R1 + R2 ) C ln 2
So
P28.65
1
2
or
e− t (R1 + R2 )C =
and
T = t1 + t 2 =
( R1 + 2 R2 ) C ln 2
A certain quantity of energy ∆Eint = P ( time ) is required to raise the temperature of the
water to100°C. For the power delivered to the heaters we have
P = I ∆V =
( ∆V )2
R
where ( ∆V ) is a constant. Thus, comparing coils 1 and 2, we have for the energy
( ∆V )2 ∆t
R1
(a)
=
( ∆V )2 2 ∆t
R2
. Then R2 = 2 R1 .
When connected in parallel, the coils present equivalent resistance
( ∆V )2 ∆t ( ∆V ) ∆t p
2R
1
1
=
=
= 1 . Now
R1
2 R1 3
1 R1 + 1 R2 1 R1 + 1 2 R1
3
2
Rp =
(b)
For the series connection, Rs = R1 + R2 = R1 + 2 R1 = 3 R1 and
∆t s = 3∆t
( ∆V )2 ∆t
R1
∆t p =
=
2 ∆t
3
( ∆V )2 ∆t s
3 R1
Direct Current Circuits
P28.66
(a)
We model the person’s body and street shoes
as shown. For the discharge to reach 100 V,
3 000 V
150 pF
80 pF
q ( t ) = Qe− t RC = C ∆V ( t ) = C ∆V0 e− t RC
∆V0
= e+t RC
∆V
∆V
= e− t RC
∆V0
149
5 000 MΩ
FIG. P28.66(a)
t
⎛ ∆V0 ⎞
= ln
⎝ ∆V ⎠
RC
t = RC ln
(b)
3 000 ⎞
⎛ ∆V0 ⎞
= 5 000 × 10 6 Ω ( 230 × 10 −12 F ) ln ⎛
= 3.91 s
⎝ 100 ⎠
⎝ ∆V ⎠
t = (1 × 10 6 V A ) ( 230 × 10 −12 C V ) ln 30 = 782 µs
ANSWERS TO EVEN PROBLEMS
P28.2
(a) 4.59 Ω (b) 8.16%
P28.4
(a) 50.0% (b) r = 0 (c) High efficiency. The electric company’s economic interest is to minimize internal energy production in its power lines, so that it can sell a large fraction of the energy
output of its generators to the customers. (d) High power transfer. Energy by electric transmission is so cheap compared to the sound system that she does not spend extra money to buy an
efficient amplifier.
P28.6
(a) The 120-V potential difference is applied across the series combination of the two conductors in the
extension cord and the light bulb. The potential difference across the light bulb is less than 120 V and
its power is less than 75 W. (b) We assume the bulb has constant resistance—that is, that its
temperature does not change much from the design operating point. See the solution. 73.8 W
P28.8
(a) See the solution.
P28.10
See the solution.
P28.12
470 Ω and 220 Ω
P28.14
(a) ∆V4 > ∆V3 > ∆V1 > ∆V2 (b) ∆V1 = e /3, ∆V2 = 2e /9, ∆V3 = 4e /9, ∆V4 = 2e /3
(c) I1 > I4 > I2 = I3 (d) I1 = I, I2 = I3 = I/3, I4 = 2I/3 (e) Increasing the value of resistor 3
increases the equivalent resistance of the entire circuit. The current in the battery, which
is also the current in resistor 1, therefore decreases. Then the potential difference across
resistor 1 decreases and the potential difference across the parallel combination increases.
Driven by a larger potential difference, the current in resistor 4 increases. This effect makes
the current in resistors 2 and 3 decrease. In summary, I4 increases while I1, I2, and I3 decrease.
(f) I1 = 3I/4, I2 = I3 = 0, I4 = 3I/4
P28.16
I1 = 714 mA I 2 = 1.29 A
P28.18
See the solution.
P28.20
(a) See the solution. (b) The current in the 220-Ω resistor and the 5.80-V battery is 11.0 mA
out of the positive battery pole. The current in the 370-Ω resistor is 9.13 mA. The current in
the 150-Ω resistor and the 3.10-V battery is 1.87 mA out of the negative battery pole.
(b) no
ε = 12.6 V
150
Chapter 28
P28.22
50.0 mA from a to e
P28.24
starter 171 A downward in the diagram; battery 0.283 A downward
P28.26
See the solution.
P28.28
587 kΩ
P28.30
See the solution.
P28.32
(a) 1.50 s
P28.34
145 Ω, 0.756 mA
P28.36
(a) 30.000 mA, 5.400 0 V (b) 30.167 mA, 5.381 6 V (c) 29.898 mA, 5.396 6 V (d) Both
circuits are good enough for some measurements. The circuit in part (c) gives data leading to a
value of resistance that is too high by only about 0.3%. Its value is more accurate than the value,
0.9% too low, implied by the data from the circuit in part (b).
P28.38
(a) ~10−14 A (b) Vh/2 + ~10−10 V and Vh/2 − ~10−10 V, where Vh is the potential of the live wire, ~102 V
P28.40
(a) 0.706 A (b) 2.49 W (c) Only the circuit in Figure P28.40c requires the use of Kirchhoff’s
rules for solution. In the other circuits the 5-Ω and 8-Ω resistors are still in parallel with each
other. (d) The power is lowest in Figure P28.40c. The circuits in Figures P28.40b and P28.40d
have in effect 30-V batteries driving the current.
P28.42
(a) either 3.84 Ω or 0.375 Ω
P28.44
(a) e 2/3R
(b) 3e 2/R
P28.46
(a) 4.40 Ω
(b) 32.0 W, 9.60 W, 70.4 W (c) 48.0 W
P28.48
(c) 200 µ A + (100 µ A ) e− t 1.00 s
(b) 1.00 s
Ps + P − 4PsP p
2
s
2I 2
(b) No load resistor can extract more than 17.6 W from this battery.
(c) in the parallel connection
and
Ps − Ps2 − 4PsP p
2I 2
160 Ω − 3 R
(b) 1.96 mC (c) Yes; 53.3 Ω
166 Ω + 83 R
corresponds to disconnecting the wire; 0.542 mC
P28.50
(a) 15.0 µ C
P28.52
See the solution.
P28.54
(a) 470 W (b) 1.60 mm or more
P28.56
(a) 222 µ C
P28.58
(a) 0.991 (b) 0.648 (c) The energy flows are precisely analogous to the currents in
parts (a) and (b). The ceiling has the smallest R-value of the thermal resistors in parallel, so
increasing its thermal resistance will produce the biggest reduction in the total energy flow.
P28.60
(a) 5.00 Ω
P28.62
(a) ln(e /∆V) = 0.011 8 t + 0.088 2
P28.64
( R1 + 2 R2 ) C ln 2
P28.66
(a) 3.91 s (b) 0.782 ms
(d) 14.5 mC for R = 0 (e) Yes; it
(c) 2.93 mm or more
(b) increase by 444 µ C
(b) 2.40 A
(b) 84.7 s, 8.47 mF
29
Magnetic Fields
CHAPTER OUTLINE
29.1
29.2
29.3
29.4
29.5
29.6
Magnetic Fields and Forces
Motion of a Charged Particle in a
Uniform Magnetic Field
Applications Involving Charged
Particles Moving in a Magnetic
Field
Magnetic Force Acting on a
Current-Carrying Conductor
Torque on a Current Loop in a
Uniform Magnetic Field
The Hall Effect
ANSWERS TO QUESTIONS
*Q29.1 (a) Yes, as described by F = qE.
(b) No, as described by F = qv × B
(c) Yes. (d) Yes. (e) No. The wire is uncharged.
(f) Yes. (g) Yes. (h) Yes.
*Q29.2 (i) (b). (ii) (a). Electron A has a smaller radius of
curvature, as described by qvB = mv2/r.
*Q29.3 (i) (c) (ii) (c) (iii) (c) (iv) (b) (v) (d) (vi) (b) (vii) (b)
(viii) (b)
*Q29.4 We consider the quantity | qvB sinq |, in units of
e (m/s)(T). For (a) it is 1 × 106 10−3 1 = 103. For (b) it is
1 × 106 10−3 0 = 0. For (c) 2 × 106 10−3 1 = 2 000. For
(d) 1 × 106 2 × 10−3 1 = 2 000. For (e) 1 × 106 10−3 1 = 103.
For (f) 1 × 106 10−3 0.707 = 707. The ranking is then
c = d > a = e > f > b.
*Q29.5
ˆi × (− kˆ ) = ˆj . Answer (c).
*Q29.6
Answer (c). It is not necessarily zero. If the magnetic field is parallel or antiparallel to the
velocity of the charged particle, then the particle will experience no magnetic force.
Q29.7
If they are projected in the same direction into the same magnetic field, the charges are of
opposite sign.
Q29.8
Send the particle through the uniform field and look at its path. If the path of the particle is
parabolic, then the field must be electric, as the electric field exerts a constant force on a charged
particle, independent of its velocity. If you shoot a proton through an electric field, it will feel a
constant force in the same direction as the electric field—it’s similar to throwing a ball through a
gravitational field.
If the path of the particle is helical or circular, then the field is magnetic.
If the path of the particle is straight, then observe the speed of the particle. If the particle
accelerates, then the field is electric, as a constant force on a proton with or against its motion will
make its speed change. If the speed remains constant, then the field is magnetic.
*Q29.9
Answer (d). The electrons will feel a constant electric force and a magnetic force that will
change in direction and in magnitude as their speed changes.
Q29.10
Yes. If the magnetic field is perpendicular to the plane of the loop, then it exerts no torque on the
loop.
151
152
Q29.11
Chapter 29
If you can hook a spring balance to the particle and measure the force on it in a known electric
F
field, then q = will tell you its charge. You cannot hook a spring balance to an electron.
E
Measuring the acceleration of small particles by observing their deflection in known electric and
magnetic fields can tell you the charge-to-mass ratio, but not separately the charge or mass. Both
an acceleration produced by an electric field and an acceleration caused by a magnetic
q
field depend on the properties of the particle only by being proportional to the ratio .
m
Q29.12 If the current loop feels a torque, it must be caused by a magnetic field. If the current loop feels
no torque, try a different orientation—the torque is zero if the field is along the axis of the loop.
Q29.13 The Earth’s magnetic field exerts force on a charged incoming cosmic ray,
tending to make it spiral around a magnetic field line. If the particle energy
is low enough, the spiral will be tight enough that the particle will first hit
some matter as it follows a field line down into the atmosphere or to the
surface at a high geographic latitude.
FIG. Q29.13
Q29.14 No. Changing the velocity of a particle requires an accelerating force. The
magnetic force is proportional to the speed of the particle. If the particle is not
moving, there can be no magnetic force on it.
SOLUTIONS TO PROBLEMS
Section 29.1
P29.1
Magnetic Fields and Forces
(a)
up
(b)
out of the page,
since the charge is
negative.
(c)
no deflection
(d)
into the page
FIG. P29.1
Magnetic Fields
P29.2
P29.3
At the equator, the Earth’s magnetic field is
horizontally north.
an electron has
Because
negative
charge,
is
opposite in direction
F
=
q
v
×
B
to v × B. Figures are drawn looking down.
(a)
Down × North = East, so the force is
directed West .
(b)
North × North = sin 0° = 0 : Zero deflection .
(c)
West × North = Down, so the force is directed
(d)
Southeast × North = Up, so the force is Down .
FIG. P29.2
Up .
F = ma = (1.67 × 10 −27 kg ) ( 2.00 × 1013 m s 2 ) = 3.34 × 10 −14 N = qv B sin 90°
B=
F
3.34 × 10 −14 N
=
= 2.09 × 10 −2 T
qv (1.60 × 10 −19 C) (1.00 × 10 7 m s)
The right-hand rule shows that B must be in the −y direction to yield a force
in the +x direction when v is in the z direction.
P29.4
153
(a)
FIG. P29.3
FB = qv B sin θ = (1.60 × 10 −19 C) ( 3.00 × 10 6 m s) ( 3.00 × 10 −1 T) sin 37.0°
FB = 8.67 × 10 −14 N
(b)
P29.5
P29.6
F 8.67 × 10 −14 N
=
= 5.19 × 1013 m s 2
m 1.67 × 10 −27 kg
FB = qv B sin θ
so
sin θ = 0.754
and
8.20 × 10 −13 N = (1.60 × 10 −19 C) ( 4.00 × 10 6 m s) (1.70 T) sin θ
θ = sin −1 ( 0.754 ) = 48.9° or 131°
First find the speed of the electron.
2 (1.60 × 10 −19 C) ( 2 400 J C)
2 e∆V
=
= 2.90 × 10 7 m s
−31
m
9
.
11
10
×
k
g
(
)
∆K =
1
m v 2 = e∆V = ∆U :
2
(a)
FB, max = qv B = (1.60 × 10 −19 C) ( 2.90 × 10 7 m s) (1.70 T) = 7.90 × 10 −12 N
(b)
P29.7
a=
v=
FB, min = 0 occurs when v is either parallel to or anti-parallel to B.
FB = qv × B
ˆi
ˆj kˆ
v × B = +2 −4 +1 = (12 − 2 ) ˆi + (1 + 6 ) ˆj + ( 4 + 4 ) kˆ = 10 ˆi + 7 ˆj + 8kˆ
+1 +2 −3
v × B = 10 2 + 72 + 82 = 14.6 T ⋅ m s
FB = q v × B = (1.60 × 10 −19 C ) (14.6 T ⋅ m s ) = 2.34 × 10 −18 N
154
P29.8
Chapter 29
qE = (−1.60 × 10 −19 C) ( 20.0 N C) kˆ = (−3.20 × 10 −18 N ) k̂
∑ F = qE + qv × B = ma
( −3.20 × 10 −18 N ) kˆ − 1.60 × 10 −19 C 1.20 × 10 4 m s ˆi × B = (9.11 × 10 −31 ) ( 2.00 × 1012 m s2 ) kˆ
− ( 3.20 × 10 −18 N ) kˆ − (1.92 × 10 −15 C ⋅ m s ) ˆi × B = (1.82 × 10 −18 N ) kˆ
(1.92 × 10 −15 C ⋅ m s ) ˆi × B = − (5.02 × 10 −18 N ) kˆ
(
)
The magnetic field may have any x -component . Bz = 0 and By = −2.62 mT .
Section 29.2
P29.9
(a)
Motion of a Charged Particle in a Uniform Magnetic Field
B = 50.0 × 10 −6 T ; v = 6.20 × 10 6 m s
Direction is given by the right-hand-rule: southward
FB = qv B sin θ
FB = (1.60 × 10 −19 C) ( 6.20 × 10 6 m s) ( 50.0 × 10 −6 T) sin 90.0°
= 4.96 × 10 −17 N
(b)
*P29.10 (a)
mv 2
so
r
2
−27
6
m v 2 (1.67 × 10 kg) (6.20 × 10 m s)
r=
=
= 1.29 km
4.96 × 10 −17 N
F
F=
FIG. P29.9
The horizontal velocity component of the electrons is given
by (1/2)mv x2 = | q |V.
vx =
2 qV
m
=
B
2(1.6 × 10 −19 C) 2 500 J/C
= 2.96 × 10 7 m/s
9.11 × 10 −31 kg
Its time of flight is t = x/vx = 0.35 m/2.96 × 107 m/s
= 1.18 × 10−8 s.
Its vertical deflection is y = (1/2) gt 2 = (1/2) 9.8 m/s2 (1.18 ×
10−8 s)2 = 6.84 × 10−16 m down , which is unobservably
small.
r
W
(b) The magnetic force is in the direction –north × down =
–west = east. The beam is deflected into a circular path with
radius
r=
mv
9.11 × 10 −31 kg 2.96 × 10 7 m/s
=
= 8.44 m.
q B 1.6 × 10 −19 C 20 × 10 –6 N ⋅ s/C ⋅ m
q
S
N
E
FIG. P29.10
Their path to the screen subtends at the center of curvature an angle given by
sin q = 0.35 m/8.44 m = 2.38°. Their deflection is 8.44 m(1 – cos 2.38°) = 7.26 mm east .
It does not move as a projectile, but its northward velocity component stays nearly constant,
changing from 2.96 × 107 m/s cos 0° to 2.96 × 107 m/s cos 2.38°. That is, it is constant
within 0.09%. It is a good approximation to think of it as moving on a parabola as it really
moves on a circle.
Magnetic Fields
P29.11
1
mv 2
2
mv 2
Also, qv B =
r
q ( ∆V ) =
2q ( ∆V )
m
m v m 2q ( ∆V )
r=
=
=
qB qB
m
v=
or
so
rp2 =
Therefore,
and
2 m ( ∆V )
qB 2
2 m p ( ∆V )
eB 2
(
)
(
)
rd2 =
⎛ 2 m p ( ∆V ) ⎞
2 md ( ∆V ) 2 2 m p ( ∆V )
=
= 2⎜
= 2rp2
2
2
qd B
eB
⎝ eB 2 ⎟⎠
rα2 =
⎛ 2 m p ( ∆V ) ⎞
2 mα ( ∆V ) 2 4 m p ( ∆V )
=
= 2⎜
= 2rp2
2
2
qα B
( 2e ) B
⎝ eB 2 ⎟⎠
rα = rd = 2 rp
The conclusion is:
P29.12
155
eBr
mv 2
.
and v =
m
r
The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic:
For each electron, q v B sin 90.0° =
1
1
1
mv12i + 0 = mv12f + mv22 f
2
2
2
1 ⎛ e2 B 2 R12 ⎞ 1 ⎛ e2 B 2 R22 ⎞ e2 B 2 2
K = m⎜
⎟ + m⎜
⎟=
( R1 + R22 )
2m
2 ⎝ m2 ⎠ 2 ⎝ m2 ⎠
K=
e (1.60 × 10 −19 C) ( 0.044 0 N ⋅ s C ⋅ m ) ⎡
2
2
K=
(0.010 0 m ) + (0.024 0 m ) ⎤⎦ = 115 keV
−31
⎣
2 ( 9.11 × 10 kg )
2
P29.13
(a)
(b)
P29.14
or
mv 2
R
qRB = m v
But
L = m v R = qR 2 B
Therefore,
R=
Thus,
v=
1
m v 2 = q ( ∆V )
2
r=
qv B =
We begin with
mv
qB
so
v=
so
r=
L
=
qB
4.00 × 10 −25 J ⋅ s
= 0.050 0 m = 5.00 cm
(1.60 × 10 −19 C) (1.00 × 10 −3 T )
L
4.00 × 10 −25 J ⋅ s
=
= 8.78 × 10 6 m s
mR ( 9.11 × 10 −31 kg ) ( 0.050 0 m )
2q ( ∆V )
m
m 2q ( ∆V ) m
qB
r2 =
m 2 ( ∆V )
⋅
q
B2
and
m′ 2 ( ∆V )
2
( r′) = ⋅ 2
m=
qB 2 r 2
2 ( ∆V )
and
( m′) =
q′
B
( q′) B 2 ( r′)
2 ( ∆V )
2
m′ q′ ( r′) ⎛ 2e ⎞ ⎛ 2 R ⎞
= ⋅
= ⎜ ⎟⎜ ⎟ = 8
⎝ e ⎠⎝ R ⎠
m q r2
2
so
2
156
P29.15
Chapter 29
E=
1
m v 2 = e∆V
2
ev B sin 90° =
and
P29.16
(a)
mv 2
R
B=
m v m 2 e∆V 1 2 m∆V
=
=
eR eR
m
R
e
B=
2 (1.67 × 10 −27 kg ) (10.0 × 10 6 V)
1
= 7.88 × 10 −12 T
5.80 × 1010 m
1.60 × 10 −19 C
The boundary between a region of strong magnetic field and a region
of zero field cannot be perfectly sharp, but we ignore the thickness
of the transition zone. In the field the electron moves on an arc of a
circle:
∑ F = ma:
q v B sin 90° =
q B
v
=ω =
=
r
m
mv 2
r
(1.60 × 10−19 C) (10−3 N ⋅ s C ⋅ m )
(9.11 × 10
−31
kg )
FIG. P29.16(a)
= 1.76 × 10 8 rad s
The time for one half revolution is,
∆θ = ω ∆ t
π rad
∆θ
∆t =
=
= 1.79 × 10 −8 s
ω 1.76 × 108 rad s
from
(b)
The maximum depth of penetration is the radius of the path.
v = ω r = (1.76 × 108 s−1 ) ( 0.02 m ) = 3.51 × 10 6 m s
Then
and
K=
2
1
1
5.62 × 10 −18 J ⋅ e
m v 2 = ( 9.11 × 10 −31 kg ) ( 3.51 × 10 6 m s ) = 5.62 × 10 −18 J =
2
2
1.60 × 10 −19 C
= 35.1 eV
Section 29.3
P29.17
Applications Involving Charged Particles Moving in a Magnetic Field
FB = Fe
so
qv B = qE
where
v=
2K
and K is kinetic energy of the electron.
m
E = vB =
2 ( 750 ) (1.60 × 10 −19 )
2K
B=
( 0.015 0 ) = 244 kV m
m
9.11 × 10 −31
Magnetic Fields
P29.18
K=
1
m v 2 = q ( ∆V )
2
so
mv 2
FB = qv × B =
r
2q ( ∆V )
m
m v m 2q ( ∆V ) m 1 2 m ( ∆V )
=
=
qB q
B
B
q
2 ( 238 × 1.66 × 10 −27 ) 2 000 ⎛ 1 ⎞
⎜
⎟ = 8.28 × 10 −2 m = 8.28 cm
⎝ 1.20 ⎠
1.60 × 10 −19
(a)
r238 =
(b)
r235 = 8.23 cm
r238
=
r235
r=
v=
m238
=
m235
238.05
= 1.006 4
235.04
The ratios of the orbit radius for different ions are independent of ∆V and B.
P29.19
P29.20
In the velocity selector:
v=
E 2 500 V m
=
= 7.14 × 10 4 m s
B 0.035 0 T
In the deflection chamber:
r=
−26
4
m v ( 2.18 × 10 kg ) ( 7.14 × 10 m s )
=
= 0.278 m
qB
(1.60 × 10 −19 C) ( 0.035 0 T )
Note that the “cyclotron frequency” is an angular speed. The motion of the proton is
described by
∑ F = ma:
q v B sin 90° =
q B=m
(a)
(b)
(c)
(d)
mv 2
r
v
= mω
r
(1.60 × 10 C) (0.8 N ⋅ s C ⋅ m ) ⎛⎜ kg ⋅ m ⎞⎟ = 7.66 × 10
⎝ N⋅s ⎠
m
(1.67 × 10 kg)
⎛ 1 ⎞
v = ω r = ( 7.66 × 10 rad s) ( 0.350 m ) ⎜
⎟ = 2.68 × 10 m s
⎝ 1 rad ⎠
ω=
q B
−19
=
−27
7
2
7
7
2⎛
1
1
1 eV ⎞
K = m v 2 = (1.67 × 10 −27 kg) ( 2.68 × 10 7 m s) ⎜
⎟ = 3.76 × 10 6 eV
⎝ 1.66 × 10 −19 J ⎠
2
2
The proton gains 600 eV twice during each revolution, so the number of revolutions is
3.76 × 10 6 eV
= 3.13 × 10 3 revolutions
2 ( 600 eV )
(e)
P29.21
(a)
(b)
rad s
θ =ωt
t=
θ
3.13 × 10 3 rev ⎛ 2π rad ⎞
=
⎟ = 2.57 × 10 −4 s
⎜
ω 7.66 × 10 7 rad s ⎝ 1 rev ⎠
mv 2
R
−19
v qBR qB (1.60 × 10 C) ( 0.450 T)
ω= =
=
=
= 4.31 × 10 7 rad s
R mR
m
1.67 × 10 −27 kg
FB = qv B =
v=
−19
qBR (1.60 × 10 C) ( 0.450 T)(1.20 m )
=
= 5.17 × 10 7 m s
m
1.67 × 10 −27 kg
157
158
Chapter 29
*P29.22 (a) The path radius is r = mv/qB, which we can put in terms of energy E by (1/2)mv2 = E.
v = (2E/m)1/2 so
r = (m/qB) (2E/m)1/2 = (2m)1/2(qB)−1E1/2
Then dr/dt = = (2m)1/2(qB)−1(1/2) E −1/2 dE/dt =
2 m 2 m q 2 B∆V 1 ∆V
=
qB2 qBr π m
r πB
(b) The dashed red line should spiral around many times, with its turns relatively far apart on
the inside and closer together on the outside.
P29.23
(c)
dr 1 ∆V 600 V
CmNm
=
=
= 682 m/s
dt r π B 0.35 m π 0.8 N s V C
(d)
∆r =
dr
1 ∆V 2π m 2∆Vm 2 × 600 V 1.67 × 10 −27 kg C2 m 2 N m
T=
=
=
= 55.9 µ m
0.35 m 1.6 × 10 −19 C 0.82 N 2 s 2 V C
dt
r π B qB
rqB 2
⎛ 25.0 ⎞
1.00 cm
θ = tan −1 ⎜
R=
= 1.08 cm
and
⎟ = 68.2°
⎝ 10.0 ⎠
sin 68.2°
Ignoring relativistic correction, the kinetic energy of the electrons is
1
m v 2 = q∆V
2
so
v=
2q∆V
= 1.33 × 108 m s
m
mv 2
= qv B, we find the magnetic field
R
−31
8
m v ( 9.11 × 10 kg ) (1.33 × 10 m s )
B=
= 70.1 mT
=
q R
(1.60 × 10 −19 C) (1.08 × 10 −2 m )
From Newton’s second law,
FIG. P29.23
*P29.24 (a) Yes: The constituent of the beam is present in all kinds of atoms.
(b) Yes: Everything in the beam has a single charge-to-mass ratio.
(c) In a charged macroscopic object most of the atoms are uncharged. A molecule never
has all of its atoms ionized. Any atom other than hydrogen contains neutrons and so has
more mass per charge if it is ionized than hydrogen does. The greatest charge-to-mass
ratio Thomson could expect was then for ionized hydrogen, 1.6 × 10−19 C/1.67 × 10−27 kg,
smaller than the value e/m he measured, 1.6 × 10−19 C/9.11 × 10−31 kg, by 1 836 times. The
particles in his beam could not be whole atoms, but rather must be much smaller in mass.
(d) With kinetic energy 100 eV, an electron has speed given by (1/2)mv2 = 100 eV
v=
200 ⋅ 1.6 × 10 −19 C 1 J/C
= 5.93 × 10 6 m/s. The time to travel 40 cm is
9.11 × 10 −31 kg
0.4 m/(5.93 × 106 m/s) = 6.75 × 10−8 s. If it is fired horizontally it will fall vertically by
(1/2)gt 2 = (1/2)(9.8 m/s2)( 6.75 × 10−8 s)2 = 2.23 × 10−14 m, an immeasurably small
amount. An electron with higher energy falls by a smaller amount.
Magnetic Fields
Section 29.4
P29.25
Magnetic Force Acting on a Current-Carrying Conductor
FB = ILB sin θ
with
FB = Fg = mg
mg = ILB sin θ
so
m
g = IB sin θ
L
FIG. P29.25
I = 2.00 A
⎛ 100 cm m ⎞
m
= ( 0.500 g cm ) ⎜
= 5.00 × 10 −2 kg m
⎝ 1 000 g kg ⎟⎠
L
and
(5.00 × 10 ) (9.80) = (2.00) B sin 90.0°
−2
Thus
B = 0.245 Tesla with the direction given by right-hand rule: eastward .
P29.26
FB = I × B = ( 2.40 A ) ( 0.750 m ) ˆi × (1.60 T ) kˆ =
P29.27
(a)
FB = ILB sin θ = ( 5.00 A)( 2.80 m )( 0.390 T) sin 60.0° = 4.73 N
(b)
FB = ( 5.00 A)( 2.80 m )( 0.390 T) sin 90.0° = 5.46 N
(c)
FB = ( 5.00 A)( 2.80 m )( 0.390 T) sin 120° = 4.73 N
( −2.88 ĵ) N
*P29.28 The magnetic force should counterbalance the gravitational force on each section of wire:
mg
kg 9.8 m/s 2 C m
= 2.4 × 10 −3
= 840 A
B
m 28 × 10 −6 N s
The current should be east so that the magnetic force will be east × north = up.
IB sin 90° = mg
P29.29
I=
The rod feels force FB = I d × B = Id kˆ × B − ˆj = IdB ˆi .
(
)
() ( )
()
The work-energy theorem is ( K trasn + K rot )i + ∆E = ( K trans + K rot ) f
B
d
I
1
1
0 + 0 + Fs cos θ = m v 2 + Iω 2
2
2
L
y
2
IdBL cos 0 º =
v=
v
1
1 1
3
m v 2 + ⎛ mR 2 ⎞ ⎛ ⎞ and IdBL = m v 2
⎝
⎠
⎝
⎠
2
2 2
R
4
4 IdBL
=
3m
4 ( 48.0 A ) ( 0.120 m ) ( 0.240 T ) ( 0.450 m )
= 1.07 m s
3 ( 0.720 kg )
x
z
FIG. P29.29
159
160
P29.30
Chapter 29
The rod feels force FB = I d × B = Id kˆ × B − ˆj = IdB ˆi .
() ( )
)
(
()
The work-energy theorem is ( K trans + K rot )i + ∆E = ( K trans + K rot ) f
0 + 0 + Fs cos θ =
1
1
mv 2 + Iω 2
2
2
2
IdBL cos 0 º =
P29.31
v
1
1 1
m v 2 + ⎛ mR 2 ⎞ ⎛ ⎞ and v =
⎠ ⎝ R⎠
2
2⎝2
4 IdBL
3m
The magnetic
force on each bit of ring is
Id s × B = IdsB radially inward and upward,
at angle q above the radial line. The radially inward components tend to squeeze
the ring but all cancel out as forces. The
upward components IdsB sin θ all add to
I 2 π rB sin θ up .
FIG. P29.31
*P29.32 (a)
For each segment, I = 5.00 A and B = 0.020 0 N A ⋅ m ˆj.
Segment
FB = I × B
ab
−0.400 m ˆj
0
bc
0.400 m kˆ
( 40.0 mN ) (−ˆi )
cd
da
(b)
−0.400 m ˆi + 0.400 m ˆj
0.400 m ˆi − 0.400 m kˆ
(
)
( 40.0 mN ) (−kˆ )
( 40.0 mN ) (kˆ + ˆi )
The forces on the four segments must add to zero, so the force on the fourth
segment must be the negative of the resultant of the forces on the other three.
FIG. P29.32
Magnetic Fields
P29.33
161
Take the x-axis east, the y-axis up, and the z-axis south. The field is
B = ( 52.0 µ T) cos 60.0° −kˆ + ( 52.0 µ T) sin 60.0° − ĵ
The current then has equivalent length: L ′ = 1.40 m − kˆ + 0.850 m ˆj
( )
( )
( )
()
FB = I L ′ × B = ( 0.035 0 A ) 0.850 ˆj − 1.40 kˆ m × −45.0 ˆj − 26.0 kˆ 10 −6 T
FB = 3.50 × 10 −8 N −22.1ˆi − 63.0 ˆi = 2.98 × 10 −6 N − ˆi = 2.98 µ N west
(
(
Section 29.5
P29.34
(a)
(b)
P29.35
)
)
(
)
( )
FIG. P29.33
Torque on a Current Loop in a Uniform Magnetic Field
2π r = 2.00 m
so
r = 0.318 m
2
µ = IA = (17.0 × 10 −3 A ) ⎡⎣π ( 0.318) m 2 ⎤⎦ = 5.41 mA ⋅ m 2
τ = µ × B so τ = (5.41 × 10 −3 A ⋅ m 2 ) ( 0.800 T) = 4.33 mN ⋅ m
τ = NBAI sin φ
τ = 100 ( 0.800 T ) ( 0.400 × 0.300 m 2 ) (1.20 A ) sin 60°
τ = 9.98 N ⋅ m
Note that f is the angle between the magnetic
moment and the B field. The loop will rotate so as to align the magnetic moment with the B field.
Looking down along the y-axis, the loop will
rotate in a clockwise direction.
FIG. P29.35
P29.36
Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of magnitude m Bsinq on the dipole, tending to turn the dipole moment in the direction of decreasing q.
According to the mechanical equations ∆U = – ∫dW and dW = t dq , the potential energy of the
dipole-field system is given by
θ
U−0=
∫
θ
µ B sin θ dθ = µ B ( − cos θ ) 90.0 º = − µ B cos θ + 0
or
U = −µ ⋅ B .
90.0 º
P29.37
(a)
The field exerts torque on the needle tending to align it with the field, so the minimum
energy orientation of the needle is:
pointing north at 48.0° below the horizontal
where its energy is U min = −µ B cos 0° = − ( 9.70 × 10 −3 A ⋅ m 2 ) ( 55.0 × 10 −6 T) = −5.34 × 10 −7 J.
It has maximum energy when pointing in the opposite direction,
south at 48.0° above the horizontal
where its energy is U max = −µ B cos 180° = + ( 9.70 × 10 −3 A ⋅ m 2 ) ( 55.0 × 10 −6 T) = +5.34 × 10 −7 J.
(b)
U min + W = U max:
W = U max − U min = + 5.34 × 10 −7 J − (−5.34 × 10 −7 J ) = 1.07 µ J
162
P29.38
Chapter 29
(a)
τ = µ × B,
τ = µ × B = µ B sin θ = NIAB sin θ
so
2
τ max = NIAB sin 90.0° = 1( 5.00 A) ⎡⎣π ( 0.050 0 m ) ⎤⎦ ( 3.00 × 10 −3 T) = 118 µ N ⋅ m
(b)
U = − µ ⋅ B,
so
−µB ≤ U ≤ +µB
2
Since µ B = ( NIA ) B = 1( 5.00 A ) ⎡⎣π ( 0.050 0 m ) ⎤⎦ ( 3.00 × 10 −3 T ) = 118 µ J,
the range of the potential energy is: −118 µ J ≤ U ≤ +118 µ J .
*P29.39 For a single-turn circle, r = 1.5 m/ 2p. The magnetic moment is
m = N I A = 1 (30 × 10–3 A) p (1.5 m/2p )2 = 5.37 × 10–3 A ⋅ m2
For a single-turn square, = 1.5 m/4. The magnetic moment is
m = N I A = 1 (30 × 10–3 A) (1.5 m/4)2 = 4.22 × 10–3 A ⋅ m2
For a single-turn triangle, = 1.5 m/3 = 0.5 m. The magnetic moment is
m = N I A = 1 (30 × 10–3 A) (1/2) 0.5 m (0.52 – 0.252)1/2 m = 3.25 × 10–3 A ⋅ m2
For a flat compact circular coil with N turns, r = 1.5 m/ 2pN. The magnetic moment is
m = N I A = N (30 × 10–3 A) p (1.5 m/2pN)2 = 5.37 × 10–3 A ⋅ m2/N
The magnetic moment cannot go to infinity. Its maximum value is 5.37 mA⋅m2 for a single-turn
circle. Smaller by 21% and by 40% are the magnetic moments for the single-turn square and
triangle. Circular coils with several turns have magnetic moments inversely proportional to the
number of turns, approaching zero as the number of turns goes to infinity.
P29.40
(a)
τ = µ × B = NIAB sin θ
τ max = 80 (10 −2 A ) ( 0.025 m ⋅ 0.04 m ) ( 0.8 N A ⋅ m ) sin 90° = 6.40 × 10 −4 N ⋅ m
(b)
2π rad ⎞ ⎛ 1 min ⎞
Pmax = τ maxω = 6.40 × 10 −4 N ⋅ m ( 3 600 rev min ) ⎛⎝
= 0.241 W
1 rev ⎠ ⎝ 60 s ⎠
(c)
In one half revolution the work is
W = U max − U min = −µ B cos 180° − (−µ B cos 0°) = 2µ B
= 2 NIAB = 2 ( 6.40 × 10 −4 N ⋅ m ) = 1.28 × 10 −3 J
In one full revolution, W = 2 (1.28 × 10 −3 J) = 2.56 × 10 −3 J .
(d)
Pavg =
W 2.56 × 10 −3 J
=
= 0.154 W
∆t
(1/60 ) s
The peak power in (b) is greater by the factor
π
.
2
Magnetic Fields
Section 29.6
P29.41
B=
163
The Hall Effect
28
−3
−19
−3
−12
nqt ( ∆VH ) (8.46 × 10 m ) (1.60 × 10 C) (5.00 × 10 m ) (5.10 × 10 V )
=
8.00 A
I
B = 4.31 × 10 −5 T = 43.1 µ T
P29.42
(a)
∆VH =
IB
nqt
nqt
B
0.080 0 T
=
=
= 1.14 × 10 5 T V
I
∆VH 0.700 × 10 −6 V
so
nqt ⎞
B=⎛
( ∆VH )
⎝ I ⎠
Then, the unknown field is
B = (1.14 × 10 5 T V ) ( 0.330 × 10 −6 V ) = 0.037 7 T = 37.7 mT
T
(b)
nqt
= 1.14 × 10 5 T V
I
n = (1.14 × 10 5 T V )
n = (1.14 × 10 5 T V )
so
I
qt
0.120 A
= 4.29 × 10 25 m −3
(1.60 × 10 C) ( 2.00 × 10 −3 m )
−19
Additional Problems
*P29.43 (a)
Define vector h to have
the downward direction of the
current, and vector L to be along the pipe into the page as
shown. The electric current experiences a magnetic forcce .
J
I h × B in the direction of L
(
(b)
)
The sodium, consisting of ions and electrons, flows along the
pipe transporting no net charge. But inside the section of
length L, electrons drift upward to constitute downward
electric current J × ( area ) = J Lw .
The current then feels a magnetic force I h × B = JLwhB sin 90°.
FIG. P29.43
This force along the pipe axis will make the fluid move, exerting pressure
F
JLwhB
=
= JLB
area
hw
(c)
Charge moves within the fluid inside the length L, but charge does not accumulate: the fluid
is not charged after it leaves the pump. It is not current-carrying and it is not magnetized.
164
P29.44
Chapter 29
(a)
At the moment shown in Figure 29.10, the particle must be
moving upward in order for the magnetic force on it to be
v
+
into the page, toward the center of this turn of its
B
spiral path. Throughout its motion it circulates clockwise.
FIG. P29.44(a)
(b)
After the particle has passed the middle of the bottle and moves
into the region of increasing magnetic field, the magnetic force
on it has a component to the left (as well as a radially inward
component) as shown. This force in the –x direction slows and
reverses the particle’s motion along the axis.
v
B
F
FIG. P29.44(b)
(c)
The magnetic force is perpendicular to the velocity and does no work on the particle. The
particle keeps constant kinetic energy. As its axial velocity component decreases, its
tangential velocity component increases.
(d)
The orbiting particle constitutes a loop of current in the yz
q
plane and therefore a magnetic dipole moment IA = A
T
in the –x direction. It is like a little bar magnet with its
N pole on the left.
(e)
Problem 31 showed that a nonuniform magnetic field exerts
a net force on a magnetic dipole. When the dipole is aligned
opposite to the external field, the force pushes it out of the
region of stronger field. Here it is to the left, a force of
repulsion of one magnetic south pole on another south pole.
+
N
FIG. P29.44(d)
B
N
S
S
FIG. P29.44(e)
*P29.45 The particle moves in an arc of a circle with radius
m v 1.67 × 10 −27 kg 3 × 10 7 m/s C m
= 12.5 km . It will not hit the Earth, but perform
=
1.6 × 10 −19 C 25 × 10 –6 N s
qB
a hairpin turn and go back parallel to its original direction.
r=
P29.46
∑F
∑F
y
= 0:
+ n − mg = 0
x
= 0:
−µ k n + IBd sin 90.0° = 0
B=
2
µ k mg 0.100 ( 0.200 kg ) ( 9.80 ms )
=
= 39.2 mT
Id
(10.0 A) (0.500 m )
S
Magnetic Fields
P29.47
165
The magnetic force on each proton, FB = qv × B = qv B sin 90° downward
perpendicular to velocity, causes centripetal acceleration, guiding it into
a circular path of radius r, with
mv 2
r
mv
r=
qB
qv B =
and
We compute this radius by first finding the proton’s speed:
K=
v=
Now,
(b)
1
mv 2
2
2K
=
m
r=
FIG. P29.47
2 ( 5.00 × 10 6 eV ) (1.60 × 10 −19 J eV )
1.67 × 10 −27 kg
= 3.10 × 10 7 m s
1.67 × 10 −27 kg ) ( 3.10 × 10 7 m s )
(
mv
= 6.46 m
=
qB (1.60 × 10 −19 C ) ( 0.050 0 N ⋅ s C ⋅ m )
We can most conveniently do part (b) first. From the figure observe that
1.00 m
1m
=
r
6.46 m
α = 8.90°
sin α =
(a)
The magnitude of the proton momentum stays constant, and its final y component is
− (1.67 × 10 −27 kg ) ( 3.10 × 10 7 m s) sin 8.90° = −8.00 × 10 −21 kg ⋅ m s
*P29.48 (a)
If B = Bx ˆi + By ˆj + Bz kˆ ,
FB = qv × B = e vi ˆi × Bx ˆi + By ˆj + Bz kˆ = 0 + evi By kˆ − evi Bz ˆj.
( ) (
Since the force actually experienced is FB = Fi ˆj , observe that
Bx could have any value , By = 0 ,
Bz = −
and
)
Fi
.
evi
⎛
F
FB = qv × B = e − vi ˆi × ⎜ Bx ˆi + 0 ˆj − i
evi
⎝
(b)
If v = −vi ˆi , then
(c)
If q = − e and v = −vi ˆi , then
)
(
⎞
kˆ ⎟ = − Fi ĵ
⎠
⎛
F
FB = qv × B = − e − vi ˆi × ⎜ Bx ˆi + 0 ˆj − i
evi
⎝
(
)
⎞
kˆ ⎟ = +Fi ĵ
⎠
Reversing either the velocity or the sign of the charge reverses the force.
P29.49
(a)
The net force is the Lorentz force given by
F = qE + qv × B = q E + v × B
F = ( 3.20 × 10 −19 ) ⎡ 4 ˆi − 1ˆj − 2kˆ + 2 ˆi + 3ˆj − 1kˆ × 2 ˆi + 4 ˆj + 1kˆ ⎤ N
⎣
⎦
(
)
(
) (
) (
Carrying out the indicated operations, we find:
F=
(b)
(3.52ˆi − 1.60 ˆj) × 10
⎛
⎛ Fx ⎞
−1 ⎜
θ = cos ⎜ ⎟ = cos
⎜
⎝F⎠
⎝
−1
−18
N .
3.52
( 3.52) + (1.60)
2
2
⎞
⎟ = 24.4°
⎟
⎠
)
166
Chapter 29
*P29.50 (a) The field should be in the +z direction, perpendicular to the final as well as to the initial
velocity, and with ˆi × kˆ = − ˆj as the direction of the initial force.
(b)
r=
−27
6
m v (1.67 × 10 kg ) ( 20 × 10 m s )
= 0.696 m
=
qB (1.60 × 10 −19 C ) ( 0.3 N ⋅ s C ⋅ m )
(c) The path is a quarter circle, of length (p /2)0.696 m = 1.09 m
(d) ∆t = 1.09 m/20 × 106 m/s = 54.7 ns
P29.51
A key to solving this problem is that reducing the normal force will reduce
F
the friction force: FB = BIL or B = B .
IL
I
When the wire is just able to move,
∑F
so
n = mg − FB cos θ
and
f = µ ( mg − FB cos θ )
Also,
∑F
so FB sin θ = f :
FB sin θ = µ ( mg − FB cos θ ) and FB =
We minimize B by minimizing FB:
dFB
cos θ − µ sin θ
= ( µ mg )
= 0 ⇒ µ sin θ = cos θ
dθ
(sin θ + µ cosθ )2
y
x
= n + FB cos θ − mg = 0
FIG. P29.51
= FB sin θ − f = 0
µ mg
sin θ + µ cos θ
⎛1⎞
Thus, θ = tan −1 ⎜ ⎟ = tan −1 ( 5.00 ) = 78.7° for the smallest field, and
⎝µ ⎠
(m L)
FB ⎛ µ g ⎞
=⎜ ⎟
IL ⎝ I ⎠ sin θ + µ cos θ
⎡ ( 0.200 ) ( 9..80 m s 2 ) ⎤
0.100 kg m
⎥
= 0.128 T
Bmin = ⎢
1.50 A
⎢⎣
⎥⎦ sin 78.7° + ( 0.200 ) cos 78.7°
B=
Bmin = 0.128 T pointing north at an angle of 78.7° below the horizontal
P29.52
Let vi represent the original speed of the alpha particle. Let vα and v p represent the particles’
speeds after the collision. We have conservation of momentum 4 m p vi = 4 m p vα + m p v p and the
relative velocity equation vi − 0 = v p − vα . Eliminating vi ,
4 v p − 4 vα = 4 vα + v p
3
vα = v p
8
3v p = 8 vα
For the proton’s motion in the magnetic field,
∑ F = ma
ev p B sin 90° =
m p v 2p
R
eBR
= vp
mp
For the alpha particle,
2evα B sin 90° =
4 m p vα2
rα
rα =
2 m p vα
eB
rα =
2m p 3
2 m p 3 eBR
3
vp =
R
=
eB 8
eB 8 m p
4
Magnetic Fields
P29.53
Let ∆x1 be the elongation due to the weight of the wire and
let ∆x2 be the additional elongation of the springs when the
magnetic field is turned on. Then Fmagnetic = 2 k ∆x2 where k is
the force constant of the spring and can be determined from
mg
k=
. (The factor 2 is included in the two previous
2 ∆x1
equations since there are 2 springs in parallel.) Combining
these two equations, we find
⎛ mg ⎞
mg∆x2
Fmagnetic = 2 ⎜
∆x2 =
;
∆x1
⎝ 2 ∆x1 ⎟⎠
Therefore, where I =
P29.54
but FB = I L × B = ILB
FIG. P29.53
(
)
( 0.100 )( 9.80 ) 3.00 × 10 −3
mg∆x2
24.0 V
=
= 0.588 T .
= 2.00 A, B =
IL ∆x1 ( 2.00 ) ( 0.050 0 ) 5.00 × 10 −3
12.0 Ω
(
)
Suppose the input power is
120 W = (120 V ) I :
I ~ 1 A = 10 0 A
Suppose
1 min ⎞ ⎛ 2π rad ⎞
⎟ ~ 200 rad s
⎟⎜
ω = 2 000 rev min ⎛⎜
⎝ 60 s ⎠ ⎝ 1 rev ⎠
and the output power is
20 W = τω = τ ( 200 rad s ) τ ~ 10 −1 N ⋅ m
Suppose the area is about
(3 cm ) × ( 4 cm ) ,
Suppose that the field is
or
A ~ 10 −3 m 2
B ~ 10 −1 T
Then, the number of turns in the coil may be found from τ ≅ NIAB :
0.1 N ⋅ m ~ N (1 C s ) (10 −3 m 2 ) (10 −1 N ⋅ s C ⋅ m )
giving
P29.55
167
N ~ 10 3
µ
The sphere is in translational equilibrium; thus
fs − Mg sin θ = 0
θ
B
(1)
fs
The sphere is in rotational equilibrium. If torques are taken
about the center of the sphere, the magnetic field produces
a clockwise torque of magnitude µ B sin θ , and the frictional
force a counterclockwise torque of magnitude fs R, where R
is the radius of the sphere. Thus:
fs R − µ B sin θ = 0
(2)
I
θ
Mg
FIG. P29.55
From (1): fs = Mg sin θ . Substituting this in (2) and canceling out sin θ , one obtains
µ B = MgR.
(3)
( 0.08 kg) ( 9.80 m s2 )
Mg
=
= 0.713 A .
π NBR π ( 5) ( 0.350 T ) ( 0.2 m )
The current must be counterclockwise as seen from above.
Now µ = NI π R 2 . Thus (3) gives I =
168
P29.56
Chapter 29
Call the length of the rod L and the tension in each wire alone
∑ F = T sin θ − ILB sin 90.0° = 0
∑ F = T cosθ − mg = 0,
x
or
T sin θ = ILB
y
or
T cos θ = mg
or
B=
tan θ =
P29.57
ILB
IB
=
mg ( m L ) g
∑ F = ma or qv B sin 90.0° = mrv
(m L) g
I
T
. Then, at equilibrium:
2
tan θ =
λg
tan θ
I
2
∴ the angular frequency for each ion is
v
qB
=ω =
= 2π f and
r
m
−19
⎛ 1
1 ⎞
1 ⎞ (1.60 × 10 C) ( 2.40 T) ⎛ 1
∆ω = ω12 − ω14 = qB ⎜
−
−
⎟=
⎟
⎜
−27
⎝
12
.
0
u
14
.
0 u⎠
m
m
1
.
66
10
kg
u
×
⎝ 12
(
)
14 ⎠
∆ω = 2.75 × 10 6 s−1 = 2.75 Mrad/s
P29.58
Let vx and v⊥ be the components of the velocity of the positron
parallel to and perpendicular to the direction of the magnetic field.
(a)
The pitch of trajectory is the distance moved along x by the
positron during each period, T (determined by the cyclotron
frequency):
⎛ 2π m ⎞
p = v x T = ( v cos 85.0°) ⎜
⎟
⎝ Bq ⎠
FIG. P29.58
(5.00 × 10 ) (cos 85.0°)(2π ) (9.11 × 10 ) =
p=
0.150 (1.60 × 10 )
−31
6
−19
(b)
The equation about circular motion in a magnetic field still applies to the radius of the
m v⊥ m v sin 85.0°
r=
=
spiral:
Bq
Bq
(9.11 × 10 )(5.00 × 10 ) (sin 85.0°) =
(0.150) (1.60 × 10 )
−31
r=
P29.59
1.04 × 10 −4 m
6
−19
1.89 × 10 −4 m
τ = IAB where the effective current due to the orbiting electrons is
and the period of the motion is
The electron’s speed in its orbit is found by requiring
ke q 2 m v 2
or
=
R2
R
Substituting this expression for v into the equation for T, we find
(9.11 × 10 ) (5.29 × 10 )
(1.60 × 10 ) (8.99 × 10 )
−31
T = 2π
−19 2
∆q q
=
∆t T
2π R
T=
v
I=
v=q
ke
mR
T = 2π
−11 3
9
= 1.52 × 10 −16 s
2
⎛q⎞
1.60 × 10 −19
π (5.29 × 10 −11 ) ( 0.400 ) = 3.70 × 10 −24 N ⋅ m .
Therefore, τ = ⎜ ⎟ AB =
−16
⎝T ⎠
1.52 × 10
mR3
q 2 ke
Magnetic Fields
P29.60
(a)
1
K = m v 2 = 6.00 MeV = (6.00 × 10 6 eV ) (1.60 × 10 −19 J eV )
2
'x x x x x
Bin = 1.00 T
K = 9.60 × 10 −13 J
v=
2 (9.60 × 10 −13 J)
1.67 × 10
−27
kg
x
= 3.39 × 10 7 m s
v
2
mv
FB = qv B =
so
R
x
45°x
45°x
R
x
x
45.0°x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
FIG. P29.60
m v (1.67 × 10 kg)(3.39 × 10 m s)
=
= 0.354 m
qB
(1.60 × 10−19 C) (1.00 T)
−27
R=
169
7
Then, from the diagram, x = 2 R sin 45.0° = 2 ( 0.354 m ) sin 45.0° = 0.501 m .
P29.61
(b)
From the diagram, observe that θ ′ = 45.0° .
(a)
The magnetic force acting on ions in the blood stream
will deflect positive charges toward point A and negative
charges toward point B. This separation of charges produces
an electric field directed from A toward B. At equilibrium,
the electric force caused by this field must balance the
magnetic force, so
⎛ ∆V ⎞
qv B = qE = q ⎜
⎟
⎝ d ⎠
or
P29.62
v=
(160 × 10−6 V)
∆V
=
= 1.333 m s
Bd ( 0.040 0 T) (3.00 × 10 −3 m )
FIG. P29.61
(b)
No . Negative ions moving in the direction of v would be deflected toward point B,
giving A a higher potential than B. Positive ions moving in the direction of v would be
deflected toward A, again giving A a higher potential than B. Therefore, the sign of the
potential difference does not depend on whether the ions in the blood are positively or
negatively charged.
(a)
See graph to the right. The Hall
voltage is directly proportional to
the magnetic field. A least-square
fit to the data gives the equation
of the best fitting line as:
∆VH = (1.00 × 10 −4 V T) B
(b)
Comparing the equation of the
line which fits the data best to
⎛ 1 ⎞
∆VH = ⎜
B
⎝ nqt ⎟⎠
120
100
80
∆VH ( m V)
60
40
20
0
0
0.2
0.4
0.6
0.8
1.0
1.2
B (T)
FIG. P29.62
continued on next page
170
Chapter 29
observe that:
I
I
= 1.00 × 10⫺4 V T , or t =
.
nqt
nq (1.00 × 10⫺4 V T)
Then, if I = 0.200 A, q = 1.60 × 10⫺19 C, and n = 1.00 × 10 26 m⫺3 , the thickness of the
sample is
t=
P29.63
0.200 A
= 1.25 × 10 −4 m = 0.125 mm
(1.00 × 10 m ) (1.60 × 10⫺19 C) (1.00 × 10⫺4 V T )
26
⫺3
When in the field, the particles follow a circular path
mv 2
according to qv B =
, so the radius of the path is
r
mv
.
r=
qB
mv
qBh
(a) When r = h =
, that is, when v =
, the
qB
m
particle will cross the band of field. It will move
in a full semicircle of radius h, leaving the field at
( 2h, 0, 0 ) with velocity v f = −vˆj .
FIG. P29.63
qBh
mv
, the particle will move in a smaller semicircle of radius r =
< h.
m
qB
It will leave the field at ( 2r, 0, 0 ) with velocity v f = −v ˆj .
(b)
When v <
(c)
When v >
qBh
mv
, the particle moves in a circular arc of radius r =
> h, centered at
m
qB
( r, 0, 0 ) . The arc subtends an angle given by θ = sin −1 ⎛⎝ h ⎞⎠ . It will leave the field at the
r
point with coordinates [ r (1 − cos θ ) , h, 0 ] with velocity v = v sin θ ˆi + v cos θ ˆj .
f
P29.64
(a)
I=
⎛ ev ⎞ 2
−24
2
µ = IA = ⎜
⎟ π r = 9.27 × 10 A ⋅ m
⎝ 2π r ⎠
ev
2π r
The Bohr model predicts the correct magnetic moment. However, the
“planetary model” is seriously deficient in other regards.
(b)
*P29.65 (a)
Because the electron is (–), its [conventional] current is clockwise, as seen
FIG. P29.64
from above, and µ points downward .
The torque on the dipole τ = µ × B has magnitude mB sin q ≈ mBq, proportional to the
angular displacement if the angle is small. It is a restoring torque, tending to turn the dipole
toward its equilibrium orientation. Then the statement that its motion is simple harmonic is
true for small angular displacements.
(b)
t = Ia
d 2q/dt 2 = –(mB/I)q = –w 2q
1 µB
is the frequency
where w = ( mB/I)1/2 is the angular frequency and f = w/2p =
2π
I
in hertz.
(c)
The equilibrium orientation of the needle shows the direction of the field. In a stronger
field, the frequency is higher. The frequency is easy to measure precisely over a wide range
of values. The equation in part (c) gives
and
4.90 Hz = (1/2p) (m/I)1/2 (B2)1/2. We square
0.680 Hz = (1/2p) (m/I)1/2 (39.2 mT)1/2
and divide to find B2 /39.2 mT = (4.90 Hz/0.680 Hz)2 so B2 = 51.9(39.2 mT) = 2.04 mT .
becomes
–mBq = I d 2q/dt 2
Magnetic Fields
171
ANSWERS TO EVEN PROBLEMS
P29.2
(a) west (b) no deflection
P29.4
(a) 86.7 f N
(b) 51.9 Tm s 2
P29.6
(a) 7.90 pN
(b) 0
P29.8
By = −2.62 mT. Bz = 0. Bx may have any value.
P29.10
(a) 6.84 × 10−16 m down (b) 7.24 mm east. The beam moves on an arc of a circle rather than
on a parabola, but its northward velocity component stays constant within 0.09%, so it is a good
approximation to treat it as constant.
P29.12
115 keV
P29.14
m′
=8
m
P29.16
(a) 17.9 ns
P29.18
(a) 8.28 cm
P29.20
(a) 7.66 × 107 rad/s (b) 26.8 Mm/s
P29.22
(b) The dashed red line should spiral around many times, with its turns relatively far apart on the
inside and closer together on the outside. (c) 682 m/s (d) 55.9 mm
P29.24
(a) Yes: The constituent of the beam is present in all kinds of atoms. (b) Yes: Everything in the
beam has a single charge-to-mass ratio. (c) Thomson pointed out that ionized hydrogen had the
largest charge-to-mass ratio previously known, and that the particles in his beam had a chargeto-mass ratio about 2 000 times larger. The particles in his beam could not be whole atoms, but
rather must be much smaller in mass. (d) No. The particles move with speed on the order of ten
million meters per second, to fall by an immeasurably small amount over a distance of less than
a meter.
P29.26
( −2.88 ĵ) N
P29.28
840 A east
(c) up
(d) down
(b) 35.1 eV
(b) 8.23 cm; ratio is independent of both ∆V and B
(c) 3.76 MeV (d) 3.13 × 103 rev
(e) 257 ms
12
P29.32
⎛ 4 IdBL ⎞
⎝ 3m ⎠
(b) The forces
(a) Fab = 0; Fbc = 40.0 mN − ˆi ; Fcd = 40.0 mN − kˆ ; Fda = ( 40.0 mN ) ˆi + kˆ
on the four segments must add to zero, so the force on the fourth segment must be the negative of
the resultant of the forces on the other three.
P29.34
(a) 5.41 mA ⋅ m 2
P29.36
See the solution.
P29.38
(a) 118 mN · m
(b) −118 mJ ≤ U ≤ 118 mJ
P29.40
(a) 640 mN ⋅ m
(b) 241 mW (c) 2.56 mJ
P29.30
( )
( )
(b) 4.33 mN ⋅ m
(d) 154 mW
(
)
172
Chapter 29
(b) 4.29 × 10 25 m 3
P29.42
(a) 37.7 mT
P29.44
See the solution.
P29.46
39.2 mT
P29.48
(a) Bx is indeterminate. By = 0. Bz =
P29.50
(a) the +z direction
P29.52
3R
4
P29.54
B ~ 10 −1 T; τ ~ 10 −1 N ⋅ m; I ~ 1 A; A ~ 10 −3 m 2 ; N ~ 10 3
P29.56
λ g tan θ
I
P29.58
(a) 0.104 mm; (b) 0.189 mm
P29.60
(a) 0.501 m
P29.62
(a) See the solution. Empirically, ∆VH = (100 µ V T ) B
P28.64
(a) 9.27 × 10−24 A · m2 (b) down
(b) 0.696 m
−Fi
evi
(b) −Fi ĵ
(c) 1.09 m
(c) +Fi ĵ
(d) 54.7 ns
(b) 45.0°
(b) 0.125 mm
30
Sources of the Magnetic Field
CHAPTER OUTLINE
30.1
30.2
30.3
30.4
30.5
30.6
30.7
The Biot-Savart Law
The Magnetic Force Between
Two Parallel Conductors
Ampère’s Law
The Magnetic Field of a Solenoid
Gauss’s Law in Magnetism
Magnetism in Matter
The Magnetic Field of the Earth
ANSWERS TO QUESTIONS
*Q30.1 Answers (b) and (c).
*Q30.2 (i) Magnetic field lines line in horizontal planes and
go around the wire clockwise as seen from above. East
of the wire the field points horizontally south. Answer (b).
(ii) The same. Answer (b).
*Q30.3 (i) Answer (f). (ii) Answer (e).
Q30.4
The magnetic field created by
wire 1 at the position of wire 2 is
into the paper. Hence, the magnetic
force on wire 2 is in direction
down × into the paper = to the right,
away from wire 1. Now wire 2
creates a magnetic field into the
page at the location of wire 1, so
wire 1 feels force up × into the
paper = left, away from wire 2.
FIG. Q30.4
*Q30.5 Newton’s third law describes the relationship. Answer (c).
*Q30.6 (a) No (b) Yes, if all are alike in sign. (c) Yes, if all carry current in the same direction. (d) no
Q30.7
Ampère’s law is valid for all closed paths surrounding a conductor, but not always convenient.
There are many paths along which the integral is cumbersome to calculate, although not
impossible. Consider a circular path around but not coaxial with a long, straight currentcarrying wire.
Q30.8
The Biot-Savart law considers the contribution of each element of current in a conductor to
determine the magnetic field, while for Ampère’s law, one need only know the current passing
through a given surface. Given situations of high degrees of symmetry, Ampère’s law is more
convenient to use, even though both laws are equally valid in all situations.
Q30.9
Apply Ampère’s law to the circular path labeled 1 in the picture. Since there is
no current inside this path, the magnetic field inside the tube must be zero. On
the other hand, the current through path 2 is the current carried by the conductor.
Therefore the magnetic field outside the tube is nonzero.
FIG. Q30.9
173
174
Chapter 30
*Q30.10 (i) Answer (b).
(ii) Answer (d), according to B =
µ0 NI
.
(iii) Answer (b).
(iv) Answer (c).
*Q30.11 Answer (a). The adjacent wires carry currents in the same direction.
*Q30.12 Answer (c). The magnetic flux is Φ B = BA cos θ . Therefore the flux is maximum when the field
is perpendicular to the area of the loop of wire. The flux is zero when there is no component of
magnetic field perpendicular to the loop—that is, when the plane of the loop contains the x axis.
*Q30.13 Zero in each case. The fields have no component perpendicular to the area.
*Q30.14 (a) Positive charge for attraction. (b) Larger. The contributions away from + and toward – are in
the same direction at the midpoint. (c) Downward (d) Smaller. Clockwise around the left-hand
wire and clockwise around the right-hand wire are in opposite directions at the midpoint.
*Q30.15 (a)
In units of m0(ampere/cm), the field of the straight wire, from m0I/2p r, is 3/(2p 2) = 0.75/p.
As a multiple of the same quantity, Nm0I/2r gives for (b) 10 × 0.3/2 × 2 = 0.75.
(c)
Nm0I/ gives 1000 × 0.3/200 = 1.5 times m0(ampere/cm), which is also
1.5 × 4p × 10–7/0.01 T = 0.19 mT.
(d)
The field is zero at the center.
(e)
1 mT is larger than 0.19 mT, so it is largest of all.
The ranking is then e > c > b > a > d.
*Q30.16 Yes. Either pole of the magnet creates field that turns atoms inside the iron to align their
magnetic moments with the external field. Then the nonuniform field exerts a net force on each
atom toward the direction in which the field is getting stronger.
A magnet on a refrigerator door goes through the same steps to exert a strong normal force on the
door. Then the magnet is supported by a frictional force.
Q30.17 Magnetic domain alignment within the magnet and then within the first piece of iron creates an
external magnetic field. The field of the first piece of iron in turn can align domains in another
iron sample. A nonuniform magnetic field exerts a net force of attraction on magnetic dipoles
aligned with the field.
Q30.18 The shock misaligns the domains. Heating will also decrease magnetism.
Q30.19
The north magnetic pole is off the coast of Antarctica, near the south geographic pole. Straight up.
*Q30.20 (a)
The third magnet from the top repels the second one with a force equal to the weight of the top
two. The yellow magnet repels the blue one with a force equal to the weight of the blue one.
(b)
The rods (or a pencil) prevents motion to the side and prevents the magnets from rotating
under their mutual torques. Its constraint changes unstable equilibrium into stable.
(c)
Most likely, the disks are magnetized perpendicular to their flat faces, making one face a
north pole and the other a south pole. One disk has its north pole on the top side and the
adjacent magnets have their north poles on their bottom sides.
(d)
If the blue magnet were inverted, it and the yellow one would stick firmly together. The
pair would still produce an external field and would float together above the red magnets.
Sources of the Magnetic Field
175
SOLUTIONS TO PROBLEMS
Section 30.1
The Biot-Savart Law
P30.1
B=
µ 0 I µ 0 q ( v 2π R )
=
= 12.5 T
2R
2R
P30.2
B=
−7
µ 0 I ( 4 π × 10 ) (1.00 A)
=
= 2.00 × 10 −7 T
2π r
2 π (1.00 m )
P30.3
(a)
B=
4µ 0 I ⎛
3π ⎞
π
⎜ cos − cos ⎟ where a =
⎝
⎠
4π a
4
4
2
is the distance from any side to the center.
B=
(b)
FIG. P30.3
For a single circular turn with 4 = 2π R,
B=
P30.4
4.00 × 10 −6 ⎛ 2
2⎞
−5
+
⎟ = 2 2 × 10 T = 28.3 µ T into the paper
⎜
0.200 ⎝ 2
2 ⎠
2
−7
µ 0 I µ 0 π I ( 4 π × 10 ) (10.0 )
=
=
= 24.77 µ T into the paper
2R
4
4 ( 0.400 )
We can think of the total magnetic field as the superposition of the field due to the long straight
wire (having magnitude µ 0 I and directed into the page) and the field due to the circular loop
2π R
µ0 I
(having magnitude
and directed into the page). The resultant magnetic field is:
2R
⎛ 1⎞µ I
B = ⎜1 + ⎟ 0 ( directed into the page ) .
⎝ π ⎠ 2R
P30.5
For leg 1, d s × rˆ = 0, so there is no contribution to the field from
this segment. For leg 2, the wire is only semi-infinite; thus,
µ0 I
1⎛ µ I ⎞
B= ⎜ 0 ⎟ =
into the paper
2 ⎝ 2π x ⎠
4π x
FIG. P30.5
176
P30.6
Chapter 30
Along the axis of a circular loop of radius R,
B=
µ 0 IR 2
2 ( x 2 + R2 )
32
⎤
B ⎡
1
⎥
=⎢
B0 ⎢⎣ ( x R )2 + 1⎥⎦
32
or
where
B0 ≡
µ0 I
2R
FIG. P30.6
xR
0.00
1.00
2.00
3.00
4.00
5.00
P30.7
B B0
1.00
0.354
0.0894
0.0316
0.0143
0.00754
Wire 1 creates at the origin magnetic field
µI
µ I
B1 = 0 right hand rule = 0 1
2π r
2π a
(a)
If the total field at the origin is
=
µ 0 I1
ĵ
2π a
2µ 0 I1 ˆ µ 0 I1 ˆ j=
j + B2 then the second wire must create field
2π a
2π a
µ I
µ0 I 2
according to B2 = 0 1 ˆj =
2π a
2π ( 2a )
.
Then I 2 = 2 I1 out of the paper .
(b)
2µ I
µ I
The other possibility is B1 + B2 = 0 1 − ˆj = 0 1 ˆj + B2 .
2π a
2π a
3µ I
µ0 I 2
Then B2 = 0 1 − ˆj =
I 2 = 6 I1 into the paper
2π a
2π ( 2a )
( )
( )
P30.8
Every element of current creates magnetic field in the same direction, into the page, at the center of the
arc. The upper straight portion creates one-half of the field that an infinitely long straight wire would
create. The curved portion creates one quarter of the field that a circular loop produces at its center.
1 µ0 I
The lower straight segment also creates field
.
2 2π r
The total field is
⎛ 1 µ0 I 1 µ0 I 1 µ0 I ⎞
µ I ⎛1 1⎞
B=⎜
+
+
⎟ into the page = 0 ⎜ + ⎟ into the plane of the paper
⎝ 2 2π r 4 2r 2 2π r ⎠
2r ⎝ π 4 ⎠
⎛ 0.284 15µ 0 I ⎞
=⎜
⎟ into the page
⎝
⎠
r
Sources of the Magnetic Field
P30.9
(a)
Above the pair of wires, the field out of the page of the 50 A
current will be stronger than the −k̂ field of the 30 A
current, so they cannot add to zero. Between the wires, both
produce fields into the page. They can only add to zero below
the wires, at coordinate y = − y . Here the total field is
( )
µ I
B= 0
2π r
+
µ0 I
2π r
:
µ ⎡
50 A
30 A ˆ ⎤
k⎥
0= 0 ⎢
−kˆ +
y
2 π ⎢⎣ ( y + 0.28 m )
⎥⎦
( )
FIG. P30.9
()
50 y = 30 ( y + 0.28 m )
50 (− y ) = 30 ( 0.28 m − y )
−20 y = 30 ( 0.28 m ) at y = −0.420 m
(b)
µ I
µ I
At y = 0.1 m the total field is B = 0
+ 0
:
2π r
2π r
4π × 10 −7 T ⋅ m A ⎛
⎞
50 A
30 A
− kˆ +
B=
− kˆ ⎟ = 1.16 × 10 −4 T − kˆ
⎜
2π
0.10 m
⎝ ( 0.28 − 0.10 ) m
⎠
( )
( )
( )
The force on the particle is
F = qv × B = (−2 × 10 −6 C) (150 × 10 6 m s) ˆi × (1.16 × 10 −4 N ⋅ s C ⋅ m ) −kˆ
()
( )
( )
= 3.47 × 10 −2 N − ˆj
(c)
We require
So
Fe = 3.47 × 10 −2 N + ˆj = qE = ( −2 × 10 −6 C ) E
( )
E = −1.73 × 10 4 ˆj N C
*P30.10 We use the Biot-Savart law. For bits of wire along the straight-line
sections, d s is at 0° or 180° to rˆ , so d s × rˆ = 0. Thus, only the
curved section of wire contributes to B at P. Hence, d s is tangent to
the arc and r̂ is radially inward; so d s × rˆ = ds 1sin 90° ⊗ = ds ⊗ .
All points along the curve are the same distance r = 0.600 m from the
field point, so
µ I d s × rˆ
µ I
µ I
B = ∫ dB = ∫ 0
= 0 2 ∫ ds = 0 2 s
2
4
π
4
π
4
πr
r
r
all current
where s is the arc length of the curved wire,
⎛ 2π ⎞
s = rθ = ( 0.600 m ) ( 30.0°) ⎜
⎟ = 0.314 m
⎝ 360° ⎠
Then, B = (10 −7 T ⋅ m A )
(3.00 A ) 0.314 m
)
2 (
( 0.600 m )
B = 262 nT into the page
FIG. P30.10
177
178
Chapter 30
*P30.11 Label the wires 1, 2, and 3 as shown in Figure (a) and let the
magnetic field created by the currents in these wires be B1 ,
B2 , and B3 , respectively.
µ0 I
µ0 I
and B3 =
.
(a) At point A: B1 = B2 =
2π (3a )
2π a 2
(
)
The directions of these fields are shown in Figure
(b).
Observe that the horizontal components of B1 and B2 cancel while their vertical components both add onto B3 .
Therefore, the net field at point A is:
Figure (a)
BA = B1 cos 45.0° + B2 cos 45.0° + B3
=
BA
(b)
µ0 I ⎡ 2
1⎤
cos 45.0° + ⎥
2π a ⎢⎣ 2
3⎦
(4π × 10
=
T ⋅ m A ) ( 2.00 A ) ⎡ 2
1⎤
cos 45° + ⎥
⎢
−2
⎣
2π (1.00 × 10 m )
3⎦
2
−7
BA = 53.3 µ T toward the bottom of the page
At point B: B1 and B2 cancel, leaving
µ0 I
BB = B3 =
2π ( 2a )
BB =
(4π × 10
−7
T ⋅ m A ) ( 2.00 A )
2π ( 2) (1.00 × 10 −2 m )
Figure (c)
= 20.0 µ T toward the bottom of the page
(c)
FIG. P30.11
µ0 I
µ I
and B3 = 0 with the directions shown in
2π a
2π a 2
Figure (c). Again, the horizontal
components of B1 and B2 cancel. The vertical
components both oppose B3 giving
At point C: B1 = B2 =
(
)
⎡
⎤
⎤
µ0 I
µ I
µ I ⎡ 2
BC = 2 ⎢
cos 45.0°⎥ − 0 = 0 ⎢
cos 45.0° − 1⎥ = 0
⎢ 2π a 2
⎥ 2π a 2π a ⎣ 2
⎦
⎣
⎦
The upward lightning current creates field in counterclockwise horizontal circles.
µ0 I
4π × 10 −7 T m 20 × 10 3 A
righthand rule =
north = 8.00 × 10 –5 T north
2π r
2π A 50 m
(
*P30.12 (a)
Figure (b)
)
20 kA
F
21.4 mm
v
N
50 m
W
E
S
FIG. P30.12
continued on next page
B
Sources of the Magnetic Field
179
The force on the electron is
F = qv × B = −1.6 × 10 −19 C (300 m/s west) × 8 × 10 −55 (N s/C m) north
= −3.84 × 10 −21 N down siin 90° = 3.84 × 10 −21 N up
(b)
(c)
9.11 × 10 −31 kg 300 m/s
mv
=
= 2.14 × 10 −5 m . This distance is negligible
qB 1.6 × 10 –19 C 8 × 10 –5 N s/C m
compared to 50 m, so the electron does move in a uniform field.
r=
w = qB/m = 2pN/t
N=
*P30.13 (a)
qBt 1.6 × 10
=
2π m
–19
C ( 8 × 10 –5 N s/C m ) 60 × 10 –66 s
2 π 9.11 × 10 −31 kg
= 134 rev
We use Equation 30.4 in the chapter text for the field created by
a straight wire of limited length. The sines of the angles appearing in that equation are equal to the cosines of the complementary
angles shown in our diagram. For the distance a from the wire to
a
the field point we have tan 30° =
, a = 0.288 7 L. One wire
L 2
contributes to the field at P
µ I
µ0 I
B = 0 ( cos θ1 − cos θ 2 ) =
(cos 30° − cos150°)
4π a
4 π ( 0.288 7 L )
q2
L
2 q1
µ 0 I (1.732 )
1.50µ 0 I
=
=
4 π ( 0.288 7 L )
πL
a
P
I
FIG. P30.13(a)
Each side contributes the same amount of field in the same direction,
which is perpendicularly into the paper in the picture. So the total field is
⎛ 1.50µ 0 I ⎞
4.50µ 0 I
.
3⎜
⎟=
πL
⎝ πL ⎠
(b)
As we showed in part (a), one whole side of the triangle creates
µ 0 I (1.732)
. Now one-half of one
4π a
nearby side of the triangle will be half as far away from
point Pb and have a geometrically similar situation. Then it
µ I 1.732) 2µ 0 I (1.732)
creates at Pb field 0 (
. The two
=
4 π ( a 2)
4π a
half-sides shown crosshatched in the picture create at Pb field
⎛ 2µ I (1.732 ) ⎞
4 µ 0 I (1.732 ) 6µ 0 I
. The rest of the triangle
2⎜ 0
=
⎟=
π
a
π ( 0.288 7 L ) π L
4
4
⎝
⎠
field at the center
will contribute somewhat more field in the same direction,
so we already have a proof that the field at Pb is stronger .
Pb
a
FIG. P30.13(b)
180
P30.14
Chapter 30
Apply the equation from the chapter text for the field created by a straight wire of
limited length, three times:
µ I ⎛
⎞
⎞
µ I ⎛
d
a
a
+
B = 0 ⎜ cos 0 −
⎟ away from you
⎟ toward you + 0 ⎜ 2
2
2
2
2
2
4π a ⎝
4π d ⎝ d + a
d +a ⎠
d +a ⎠
+
⎞
µ 0 I ⎛ −d
− cos 180° ⎟ toward you
⎜ 2
4π a ⎝ d + a2
⎠
(
µ0 I a 2 + d 2 − d a 2 + d 2
B=
2π ad a 2 + d 2
P30.15
) away from you
Take the x-direction to the right and the y-direction up in the plane of the
paper. Current 1 creates at P a field
−7
µ I ( 2.00 × 10 T ⋅ m ) (3.00 A )
B1 = 0 =
2π a
A ( 0.050 0 m )
B1 = 12.0 µ T downward and leftward, at angle 67.4° below the –x axis.
Current 2 contributes
B2 =
(2.00 × 10
−7
T ⋅ m ) (3.00 A )
A ( 0.120 m )
clockwise perpendicular to 12.0 cm
B2 = 5.00 µ T to the right and down, at angle –22.6°
Then,
5.00 cm
I1
P
B1
13.0 cm
12.0 cm
I2
FIG. P30.15
B = B1 + B2 = (12.0 µ T) −ˆi cos 67.4° − ˆj sin 67.4° + ( 5.00 µ T) ˆi cos 22.6° − ˆj sin 22.6°
(
B = (−11.1 µ T) ˆj − (1.92 µ T) ˆj =
B2
)
(
)
(−13.0 µ T) ˆj
*P30.16 We apply B = (m 0 /2p) m/x3 to the center of a face, where B = 40 000 mT and x = 0.6 mm, and also
to the exterior weak-field point, where B = 50 m T and x is the unknown.
Then Bx3 = Bx3 40 000 mT (0.6 mm)3 = 50 mT (d + 0.6 mm)3
d = (40 000/50)1/3 0.6 mm – 0.6 mm = 4.97 mm
The strong field does not penetrate your painful joint.
Section 30.2
P30.17
The Magnetic Force Between Two Parallel Conductors
By symmetry, we note that the magnetic forces on the top and bottom
segments of the rectangle cancel. The net force on the vertical segments of
the rectangle is (using an equation from the chapter text)
µ II 1
µ I I ⎛ −a ⎞ ˆ
1
F = F1 + F2 = 0 1 2 ⎛
− ⎞ ˆi = 0 1 2 ⎜
i
2π ⎝ c + a c ⎠
2π ⎝ c ( c + a ) ⎟⎠
( 4π × 10 −7 N A 2 )( 5.00 A ) (10.0 A ) ( 0.450 m ) ⎛
⎞ ˆ
−0.150 m
i
F=
⎜
2π
⎝ ( 0.100 m ) ( 0.250 m ) ⎟⎠
F = −2.70 × 10 −5 ˆi N
(
or
)
F = 2.70 × 10 −5 N toward the left
FIG. P30.17
Sources of the Magnetic Field
P30.18
Let both wires carry current in the x direction, the first at y = 0
and the second at y = 10.0 cm.
(a)
µ I
( 4 π × 10−7 T ⋅ m A ) (5.00 A) k̂
B = 0 kˆ =
2π r
2 π ( 0.100 m )
B = 1.00 × 10 −5 T out of the page
181
y
I2 = 8.00 A
y = 10.0 cm
I1 = 5.00 A
x
z
FIG. P30.18(a)
(b)
(c)
(d)
P30.19
FB = I 2 × B = (8.00 A ) ⎡⎣(1.00 m ) ˆi × (1.00 × 10 −5 T) kˆ ⎤⎦ = (8.00 × 10 −5 N ) − ˆj
FB = 8.00 × 10 −5 N toward the first wire
( )
µ I
( 4 π × 10−7 T ⋅ m A ) (8.00 A) −kˆ = 1.60 × 10−5 T −kˆ
B = 0 −kˆ =
(
)
2π r
2 π ( 0.100 m )
B = 1.60 × 10 −5 T into the page
( )
( )
( )
FB = I1 × B = ( 5.00 A ) ⎡⎣(1.00 m ) ˆi × (1.60 × 10 −5 T ) − kˆ ⎤⎦ = (8.00 × 10 −5 N ) + ˆj
FB = 8.00 × 10 −5 N towards the second wire
( )
( )
To attract, both currents must be to the right. The attraction is described by
F = I 2 B sin 90° = I 2 So
I2 =
µ0 I
2π r
⎛
⎞
2 π ( 0.5 m )
F 2π r
⎟ = 40.0 A
= ( 320 × 10 −6 N m ) ⎜⎜
−7
⎟
µ 0 I1
4
10
N
π
×
⋅
s
C
⋅
m
A
20
(
)
(
)
⎝
⎠
FIG. P30.19
Let y represent the distance of the zero-field point below the upper wire.
Then
µ I
B= 0
2π r
+
20 ( 0.5 m − y ) = 40 y
µ0 I
2π r
0=
⎞
µ 0 ⎛ 20 A
40 A
⎜
(away ) +
( toward ) ⎟
2π ⎝ y
( 0.5 m − y)
⎠
20 ( 0.5 m ) = 60 y
y = 0.167 m below the upper wire
*P30.20 Carrying oppositely directed currents, wires 1 and 2 repel Wire 3 Wire 1
Wire 2
each other. If wire 3 were between them, it would have
to repel either 1 or 2, so the force on that wire could not
be zero. If wire 3 were to the right of wire 2, it would feel
I3
1.50 A
4.00 A
a larger force exerted by 2 than that exerted by 1, so the
total force on 3 could not be zero. Therefore wire 3 must
d
20 cm
be to the left of both other wires as shown. It must carry
downward current so that it can attract wire 2. We answer
part (b) first.
FIG. P30.20
(b)
For the equilibrium of wire 3 we have
F1 on 3 = F2 on 3
1.5 ( 20 cm + d ) = 4 d
µ 0 (1.50 A ) I 3
µ0 (4 A ) I 3
=
2π d
2π ( 20 cm + d )
30 cm
d=
= 12.0 cm to the left of wire 1
2.5
continued on next page
182
Chapter 30
(a)
Thus the situation is possible in just one way.
(c)
For the equilibrium of wire 1,
µ 0 I 3 (1.5 A ) µ 0 ( 4 A )(1.5 A )
=
2π (12 cm )
2π ( 20 cm )
I3 =
12
4 A = 2.40 A down
20
We know that wire 2 must be in equilibrium because the forces on it are equal in magnitude
to the forces that it exerts on wires 1 and 3, which are equal because they both balance the
equal-magnitude forces that 1 exerts on 3 and that 3 exerts on 1.
*P30.21 The separation between the wires is
a = 2 ( 6.00 cm ) sin 8.00° = 1.67 cm
(a)
Because the wires repel, the currents are in
opposite directions .
(b)
Because the magnetic force acts horizontally,
µ I 2
FB
= 0
= tan 8.00°
Fg 2π amg
I2 =
(c)
*P30.22 (a)
Section 30.3
P30.23
FIG. P30.21
so
I = 67.8 A
Smaller. A smaller gravitational force would be pulling down on the wires and so tending to pull the wires together. Then a smaller magnetic force is required to keep the wires
apart.
If the conductors are wires, F = I2B sin 90° and B = m0I1/2p r give F = I1 I2 m0 /2p r
r=
(b)
mg 2π a
tan 8.00°
µ 0
I1 I 2 µ 0 (10 A)2 2 × 10 −7 T m 0.5 m
= 10.0 µ m
=
2π F
1N A
For a force of ordinary size in a tabletop mechanics experiments to act on ordinary-to-large
size currents, the distance between them must be quite small, but the situation is physically
possible. If we tried to use wires with diameter 10 m m on a tabletop, they would feel the
force only momentarily after we turn on the current, until they melt. We can use wide, thin
sheets of copper, perhaps plated onto glass, and perhaps with water cooling, to have the
forces act continuously. A practical electric motor must use coils of wire with many turns.
Ampère’s Law
Each wire is distant from P by
( 0.200 m ) cos 45.0° = 0.141 m
Each wire produces a field at P of equal magnitude:
BA =
−7
µ0 I ( 2.00 × 10 T ⋅ m A ) ( 5.00 A )
=
= 7.07 µ T
2π a
( 0.141 m )
Carrying currents into the page, A produces at P a field
of 7.07 µ T to the left and down at –135º, while B
creates a field to the right and down at –45º. Carrying
currents toward you, C produces a field downward and
to the right at –45º, while D’s contribution is downward and to the left. The total field is then
4 ( 7.07 µ T) sin 45.0° = 20.0 µ T toward the bottom of the page.
FIG. P30.23
Sources of the Magnetic Field
P30.24
µ0 I
at the proton’s location. And we have
2π d
µ I
a balance between the weight of the proton and the magnetic force mg − ˆj + qv −ˆi × 0 kˆ = 0
2π d
at a distance d from the wire
Let the current I be to the right. It creates a field B =
( )
qvµ0 I (1.60 × 10
d=
=
2π mg
P30.25
183
−19
( )
C ) ( 2.30 × 10 4 m s ) ( 4π × 10 −7 T ⋅ m A ) (1.20 × 10 −6 A )
2π (1.67 × 10 −27 kg ) ( 9.80 m s 2 )
From Ampère’s law, the magnetic field at point a is given by Ba =
()
= 5.40 cm
µ0 I a
, where I a is the net
2π ra
current through the area of the circle of radius ra. In this case, I a = 1.00 A out of the page
(the current in the inner conductor), so
( 4π × 10 T ⋅ m A ) (1.00 A ) =
=
2π (1.00 × 10 m )
−7
Ba
Similarly at point b: Bb =
radius rb .
200 µ T toward top of page
−3
µ0 I b
, where I b is the net current through the area of the circle having
2 π rb
Taking out of the page as positive, I b = 1.00 A − 3.00 A = −2.00 A, or I b = 2.00 A into the page.
Therefore,
( 4π × 10 T ⋅ m A ) ( 2.00 A ) =
=
2π ( 3.00 × 10 m )
−7
Bb
P30.26
−3
µ0 I
, the field will be one-tenth as large at a ten-times larger distance: 400 cm .
2π r
(a)
In B =
(b)
µ I
µ I
B = 0 kˆ + 0 −kˆ
2 π r1
2 π r2
( )
B=
(c)
133 µ T toward bottom of page
so
4π × 10 −7 T ⋅ m ( 2.00 A ) ⎛
1
1
⎞
= 7.50 nT
−
⎜
⎝ 0.398 5 m 0.401 5 m ⎟⎠
2π A
Call r the distance from cord center to field point and 2 d = 3.00 mm the distance between
conductors.
B=
µ0 I ⎛ 1
1 ⎞ µ0 I 2 d
−
⎜
⎟=
2π ⎝ r − d r + d ⎠ 2π r 2 − d 2
7.50 × 10 −10 T = ( 2.00 × 10 −7 T ⋅ m A) ( 2.00 A)
( 3.00 × 10 m )
−3
r 2 − 2.25 × 10 −6 m 2
so
r = 1.26 m
The field of the two-conductor cord is weak to start with and falls off rapidly with distance.
(d)
The cable creates zero field at exterior points, since a loop in Ampère’s law encloses
zero total current. Shall we sell coaxial-cable power cords to people who worry about
biological damage from weak magnetic fields?
184
P30.27
Chapter 30
(a)
One wire feels force due to the field of the other ninety-nine.
−7
−2
µ 0 I 0 r ( 4 π × 10 T ⋅ m A) ( 99 )( 2.00 A) ( 0.200 × 10 m )
B=
=
2
2π R2
2 π ( 0.500 × 10 −2 m )
= 3.17 × 10 −3 T
This field points tangent to a circle of radius 0.200 cm and exerts
force F = I × B toward the center of the bundle, on the single
hundredth wire:
(b)
P30.28
P30.29
F
= IB sin θ = ( 2.00 A ) (3.17 × 10 −3 T) sin 90° = 6.34 mN
Nm
FIG. P30.27
FB
= 6.34 × 10 −3 N m inward
B ∝ r, so B is greatest at the outside of the bundle. Since each wire carries the same
current, F is greatest at the outer surface .
(a)
Binner =
−7
3
µ 0 NI ( 4 π × 10 T ⋅ m A ) (900 ) (14.0 × 10 A )
=
= 3.60 T
2π r
2π ( 0.700 m )
(b)
Bouter =
3
−7
µ 0 NI ( 2 × 10 T ⋅ m A ) (900 ) (14.0 × 10 A )
=
= 1.94 T
2π r
1.30 m
We assume the current is vertically upward.
(a)
Consider a circle of radius r slightly less than R. It encloses no current so from
养 B ⋅ d s = µ0 I inside
B (2π r ) = 0
we conclude that the magnetic field is zero .
(b)
Now let the r be barely larger than R. Ampère’s law
becomes B ( 2 π R) = µ 0 I ,
so
B=
µ0 I
2π R
The field’s direction is
(c)
FIG. P30.29(a)
tangent to the wall of the cylinder in a couunterclockwise sense .
Consider a strip of the wall of width dx and length . Its width is so
small compared to 2π R that the field at its location would be essentially
unchanged if the current in the strip were turned off.
Idx
The current it carries is I s =
up.
2π R
The force on it is
2
Idx ⎛ µ0 I ⎞ µ0 I dx ×
radially inward.
F = Is × B =
up
into
page
=
2π R ⎜⎝ 2π R ⎟⎠
4π 2 R 2
FIG. P30.29(c)
The pressure on the strip and everywhere on the cylinder is
P=
µ I 2 dx
F
= 02 2
=
A 4π R dx
µ0 I 2
inward
( 2π R )2
The pinch effect makes an effective demonstration when an aluminum can crushes itself as
it carries a large current along its length.
Sources of the Magnetic Field
From
P30.31
Use Ampère’s law,
2 π rB 2 π (1.00 × 10 ) ( 0.100 )
= 500 A
=
µ0
4 π × 10 −7
−3
养 B ⋅ d = µ I,
P30.30
185
I=
0
养 B ⋅ d s = µ I . For current density J, this becomes
0
养 B ⋅ d s = µ0 ∫ J ⋅ dA.
(a)
For r1 < R, this gives
B2 π r1 = µ 0
r1
∫ (br )(2π rdr )
and
FIG. P30.31
0
B=
µ0 br12
3
( for r1 < R or inside the cylinder )
R
(b)
When r2 > R, Ampère’s law yields
(2π r2 ) B = µ0 ∫ (br )(2π rdr ) =
0
or
*P30.32 (a)
(c)
B=
µ0 bR
3r2
3
(for r2 > R or outside the cylinder )
See Figure (a) to the right.
We choose to do part (c) before part (b). At a point on the
z axis, the contribution from each wire has magnitude
µ0 I
B=
and is perpendicular to the line
2π a 2 + z 2
from this point to the wire as shown in Figure (b). Combining
fields, the vertical components cancel while the horizontal
components add, yielding
⎛
⎞
⎛
⎞
µ0 I
µ0 Iz
z
µ0 I
=
By = 2 ⎜
sin θ ⎟ =
2
2
2
2
2
2 ⎜
2
2 ⎟
⎠ π a +z ⎝ a +z ⎠ π a +z
⎝ 2π a + z
(
By =
(b)
(d)
2 πµ 0 bR 3
,
3
4π × 10 −7 T m 8 A z
π ⎡⎣(0.03 m)2 + z 2 ⎤⎦
so
(Currents are into the
paper) Figure (a)
)
32 × 10 −7 z T ⋅ m
ˆj
B=
9 × 10 −4 m 2 + z 2
Substituting z = 0 gives zero for the field. We can see this from
cancellation of the separate fields in either diagram.
Taking the limit z → ∞ gives 1/z → 0, as we should expect.
At a distance z
above the plane
of the conductors
Figure (b)
FIG. P30.32
The condition for a maximum is:
dBy
dz
=
− µ0 Iz ( 2 z )
π (a2 + z 2 )
2
+
µ0 I
=0
π (a2 + z 2 )
or
2
2
µ0 I ( a − z )
=0
π ( a 2 + z 2 )2
Thus, along the z axis, the field is a maximum at d = a = 3.00 cm .
(e)
32 × 10 −7 0.03 m T ⋅ m ˆ
j = 53.3 ĵ m T .
The value of the maximum field is B =
9 × 10 −4 m 2 + 9 × 10 −4 m 2
186
Chapter 30
I
*P30.33 J s = . Each filament of current creates a contribution to the
total field that goes counterclockwise around that filament’s
location. Together, they create field straight up to the right of
the sheet and straight down to the left of the sheet.
From Ampère’s law applied to the suggested rectangle,
养 B ⋅ d s = µ0 I
B ⋅ 2 + 0 = µ 0 J s Therefore the field is uniform in
space, with the magnitude B =
µ0 J s
.
2
FIG. P30.33
Section 30.4
The Magnetic Field of a Solenoid
*P30.34 In the expression B = Nm0I/ for the field within a solenoid with radius much less than 20 cm, all
we want to do is increase N.
(a) Make the wire as long and thin as possible without melting when it carries the 5-A current.
Then the solenoid can have many turns.
(b) As small in radius as possible with your experiment fitting inside. Then with a smaller
circumference the wire can form a solenoid with more turns.
P30.35
P30.36
N
B = µ0 I
so
(
)
1.00 × 10 −4 T 0.400 m
B
I=
=
= 31.8 mA
µ0 n
4π × 10 −7 T ⋅ m A 1 000
(
)
Let the axis of the solenoid lie along the y–axis from y = 0 to y = . We will determine the field
at y = a . This point will be inside the solenoid if 0 < a < and outside if a < 0 or a > . We
think of solenoid as formed of rings, each of thickness dy. Now I is the symbol for the current in
⎛ N⎞
each turn of wire and the number of turns per length is ⎜ ⎟ . So the number of turns in the ring
⎝ ⎠
N
N
⎛ ⎞
⎛ ⎞
is ⎜ ⎟ dy and the current in the ring is I ring = I ⎜ ⎟ dy. Now we use the result derived in the
⎝ ⎠
⎝ ⎠
chapter text for the field created by one ring:
Bring =
µ0 I ring R 2
(
2 x 2 + R2
)
32
where x is the name of the distance from the center of the ring, at location y, to the field point
x = a − y . Each ring creates field in the same direction, along our y–axis, so the whole field of the
solenoid is
B=
∑
all rings
Bring = ∑
µ0 I ring R 2
2( x + R
2
)
2 32
=∫
0
µ0 I ( N ) dyR 2
(
2 (a − y) + R
2
2
To perform the integral we change variables to u = a − y.
B=
µ0 INR 2
2
continued on next page
a− ∫
a
(u
− du
2
+ R2
)
32
)
32
=
µ0 INR 2
2
dy
∫ 2 (a − y)
0
(
2
+ R2
)
32
Sources of the Magnetic Field
187
and then use the table of integrals in the appendix:
(a)
(b)
µ INR 2
−u
B= 0
2
2 R u 2 + R2
a− =
a
µ0 IN ⎡⎢
a
−
2
2 ⎢ a + R2
⎣
a−
( a − )2
⎤
⎥
+ R 2 ⎥⎦
If is much larger than R and a = 0,
we have B ≅
µ0 IN ⎡
− ⎤ µ0 IN
.
⎢0 − 2 ⎥ =
2 ⎣
2
⎦
This is just half the magnitude of the field deep within the solenoid. We would get the same
result by substituting a = to describe the other end.
P30.37
The field produced by the solenoid in its interior is
given by
⎛ 30.0 ⎞
B = µ 0 nI −ˆi = ( 4 π × 10 −7 T ⋅ m A ) ⎜ −2 ⎟ (15.0 A ) −ˆi
⎝ 10 m ⎠
B = − (5.65 × 10 −2 T) ˆi
( )
( )
The force exerted on side AB of the square current
loop is
(F )
B AB
(F )
B AB
= I L × B = ( 0.200 A ) ⎡( 2.00 × 10 −2 m ) ˆj × ( 5.65 × 10 −2 T ) − î ⎤
⎣
⎦
( )
= ( 2.26 × 10 −4 N ) kˆ
Similarly, each side of the square loop experiences a
force, lying in the plane of the loop, of
226 µ N directed away from the center .
FIG. P30.37
From the above result, it is seen that the net torque exerted on the square
loop by the field of the solenoid should be zero. More formally, the magnetic dipole moment of
the square loop is given by
2
µ = IA = ( 0.200 A ) ( 2.00 × 10 −2 m ) −ˆi = −80.0 µ A ⋅ m 2 î
The torque exerted on the loop is then τ = µ × B = −80.0 µ A ⋅ m 2 ˆi × −5.65 × 10 −2 Tˆi = 0
( )
(
) (
)
75 cm
= 750. We assume that the solenoid is long enough to qualify
0.1 cm
N µ0 I
so
as a long solenoid. Then the field within it (not close to the ends) is B =
(8 × 10 −3 T ) ( 0.75 m ⋅ A ) = 6.37 A
B
I=
=
N µ0
750 ( 4π × 10 −7 T ⋅ m )
*P30.38 The number of turns is N =
The resistance of the wire is
R=
−8
ρ wire (1.7 × 10 Ω ⋅ m ) 2π ( 0.05 m ) 750
=
= 5.10 Ω
2
A
π ( 0.05 × 10 −2 m )
The power delivered is
P = I ∆V = I 2 R = ( 6.37 A )2 ( 5.10 Ω ) = 207 W
The power required would be smaller if wire were wrapped in several layers.
188
Chapter 30
Section 30.5
P30.39
(a)
Gauss’s Law in Magnetism
ΦB =
∫ B ⋅ dA = B ⋅ A = (5ˆi + 4 ˆj + 3kˆ ) T ⋅ (2.50 × 10
−2
m ) î
2
Φ B = 3.12 × 10 −3 T ⋅ m 2 = 3.12 × 10 −3 Wb = 3.12 mWb
P30.40
(Φ B )total = 养 B ⋅ dA =
(a)
(Φ B )flat = B ⋅ A = Bπ R2 cos (180 − θ ) =
(b)
The net flux out of the closed surface is zero: ( Φ B )flat + ( Φ B )curved = 0.
(a)
(b)
0
for any closed surface (Gauss’s law for magnetism)
(Φ B )curved =
P30.41
(b)
− Bπ R 2 cos θ
Bπ R 2 cos θ
Φ B = B ⋅ A = BA where A is the cross-sectional area of the solenoid.
⎛ µ NI ⎞
Φ B = ⎜ 0 ⎟ ( π r 2 ) = 7.40 µ Wb
⎝ ⎠
⎛ µ NI ⎞
Φ B = B ⋅ A = BA = ⎜ 0 ⎟ ⎡⎣π r22 − r12 ⎤⎦
⎝ ⎠
(
)
⎡ ( 4 π × 10 −7 T ⋅ m A) ( 300 )(12.0 A) ⎤
2
⎥ π ⎡( 8.00 )2 − ( 4.00 )2 ⎤ (10 −3 m ) = 2.27 µ Wb
ΦB = ⎢
⎣
⎦
⎢⎣
⎥⎦
(0.300 m )
*P30.42 The field can be uniform in magnitude. Gauss’s law for magnetism implies that magnetic field
lines never start or stop. If the field is uniform in direction, the lines are parallel and their density
stays constant along any one bundle of lines. Thus the magnitude of the field has the same value
at all points along a line in the direction of the field. The magnitude of the field could vary over a
plane perpendicular to the lines, or it could be constant throughout the volume.
Section 30.6
P30.43
Magnetism in Matter
The magnetic moment of one electron is taken as one Bohr magneton m B. Let x represent the
number of electrons per atom contributing and n the number of atoms per unit volume. Then
nxm B is the magnetic moment per volume and the magnetic field (in the absence of any currents
in wires) is B = µ 0 nxµ B = 2.00 T.
Then x =
2.00 T
B
=
= 2.02
28
−3
µ 0 µ B n (8.50 × 10 m ) (9.27 × 10 −24 N ⋅ m T) ( 4 π × 10 −7 T ⋅ m A )
Sources of the Magnetic Field
Section 30.7
P30.44
P30.45
The Magnetic Field of the Earth
−7
µ0 NI ( 4π × 10 ) ( 5.00 )( 0.600 )
=
= 12.6 µ T
2R
0.300
(a)
Bh = Bcoil =
(b)
Bh = B sin φ → B =
(a)
12.6 µ T
Bh
=
= 56.0 µ T
sin φ sin 13.0°
FIG. P30.44
8.00 × 10 22 A ⋅ m 2
= 8.63 × 10 45 .
2
−24
9.27 × 10 A ⋅ m
Each iron atom has two unpaired electrons, so the number of iron atoms required is
1
8.63 × 10 45 .
2
Number of unpaired electrons =
(
)
( 4.31 × 10 atoms)( 7 900 kg
(8.50 × 10 atoms m )
45
Mass =
(b)
*P30.46 (a)
28
m3 )
3
= 4.01 × 10 20 kg
Gravitational field and magnetic field are vectors; atmospheric pressure is a scalar.
(b)
At 42° north latitude, 76° west longitude, 260 m above sea level: Gravitational field is
9.803 m/s2 down. From the Coast and Geodetic Survey, the magnetic field is 54 mT at 12°
west of geographic north and 69° below the horizontal. Atmospheric pressure is 98 kPa.
(c)
The atmosphere is held on by gravitation, but otherwise the effects are all separate. The
magnetic field could be produced by permanent magnetization of a cold iron-nickel deposit
within the Earth, so it need not be associated with present-day action of gravitation.
Additional Problems
P30.47
Consider a longitudinal filament of the strip of width dr as
shown in the sketch. The contribution to the field at point P
due to the current dI in the element dr is
µ dI
dB = 0
2π r
⎛ dr ⎞
dI = I ⎜ ⎟
⎝ w⎠
where
FIG. P30.47
b+ w µ Idr
µ0 I ⎛ w ⎞ ˆ
B = ∫ dB = ∫ 0
kˆ =
ln 1 +
k
2
2
π
πw ⎝
wr
b⎠
b
P30.48
189
B=
so
(
µ0 IR 2
2 R2 + R
)
2 32
=
µ0 I
25 2 R
I=
2 5 2 BR 2
=
µ0
I = 2.01 × 10 9 A toward the west
52
( 7.00 × 10 T) (6.37 × 10 m )
( 4 π × 10 T ⋅ m A)
−5
−7
6
190
P30.49
Chapter 30
At a point at distance x from the left end of the bar,
µ0 I 2
to
current I 2 creates magnetic field B =
2π h2 + x 2
h
the left and above the horizontal at angle q where
θ
x
tan θ = . This field exerts force on an element of the
I2
h
rod of length dx
µ0 I 2 dx
dF = I 1 × B = I 1
sin θ
2π h 2 + x 2 right hand ruule
µ0 I1 I 2 dx
x
=
into the page
2
2
2
2π h + x
h + x2
µ I I xdx
dF = 0 1 2 2 2
2π h + x
(
B
θ
x
I1
FIG. P30.49
− kˆ
)( )
The whole force is the sum of the forces on all of the elements of the bar:
F=
=
( ) ∫ 2 xdx
h +x
µ 0 I1 I 2 −kˆ
µ 0 I1 I 2 xdx
ˆ
−
k
=
∫
2
2
4π
x =0 2 π ( h + x )
( )
( ) ⎡ln (h
⎣
4π
µ 0 I1 I 2 −kˆ
2
+ 2 ) − ln h 2 ⎤⎦ =
2
2
=
( ) ln (h
4π
µ 0 I1 I 2 −kˆ
0
2
+x
2
)
0
10
−7
( ) ln ⎡⎢ (0.5 cm)
N (100 A ) ( 200 A ) −kˆ
A2
( )
2
+ (10 cm ) ⎤
⎥
2
( 0.5 cm )
⎦
⎣
2
( )
= 2 × 10 −3 N −kˆ ln 401 = 1.20 × 10 −2 N −kˆ
P30.50
Suppose you have two 100-W headlights running from a 12-V battery, with the whole
200 W
= 17 A current going through the switch 60 cm from the compass. Suppose
12 V
the dashboard contains little iron, so µ ≈ µ0 . Model the current as straight. Then,
µ0 I ( 4π × 10 )17
~10 −5 T
=
2π r
2π ( 0.6 )
−7
B=
If the local geomagnetic field is 5 × 10 −5 T , this is ~10 −1 times as large,
compass noticeably.
P30.51
On the axis of a current loop, the magnetic field is given by B =
where in this case I =
with a magnitude of
B=
P30.52
µ 0ω R 2 q
(
4π x + R
2
2 x 2 + R2
)
32
µ0 ( 20.0 )( 0.100 ) (10.0 × 10 −6 )
2
)
2 32
=
where in this case I =
R
2
µ0 IR 2
q
. The magnetic field is directed away from the center,
(2π ω )
4π ⎡⎣( 0.050 0 ) + ( 0.100 ) ⎤⎦
2
2
32
= 1.43 × 10 −10 T
On the axis of a current loop, the magnetic field is given by
when x =
(
enough to affect the
q
.
(2π ω )
B=
Therefore,
B=
then
B=
(
µ0 IR 2
2 x 2 + R2
)
32
µ 0ω R 2 q
(
4π x 2 + R 2
µ 0ω R 2 q
4π ( 45 R 2 )
32
)
32
=
µ0 qω
2.5 5π R
Sources of the Magnetic Field
P30.53
(a)
191
Use twice the equation for the field created by a current loop
Bx =
(
µ0 IR 2
2 x 2 + R2
)
32
If each coil has N turns, the field is just N times larger.
⎡
N µ0 IR 2 ⎢
1
B = Bx1 + Bx 2 =
⎢
2
2
2
⎢⎣ x + R
(
B=
(b)
1
N µ0 IR 2 ⎡⎢
2
2
⎢ x + R2
⎣
(
dB N µ0 IR 2
=
dx
2
)
32
+
(
Substituting x =
32
+
⎡( R − x )2
⎣
⎤
⎥
32 ⎥
2⎤
+R ⎦ ⎥
⎦
FIG. P30.53
⎤
⎥
32
2 R 2 + x 2 − 2 xR ⎥⎦
1
⎡ 3
2
2
⎢⎣− 2 ( 2 x ) x + R
(
)
1
)
−5 2
−
)
(
3
2 R 2 + x 2 − 2 xR
2
) (2 x − 2 R )⎤⎥⎦
−5 2
dB
R
=0 .
and canceling terms,
2
dx
−5 2
−7 2
−5 2
d 2 B −3N µ 0 IR 2 ⎡ 2
=
x + R 2 ) − 5 x 2 ( x 2 + R 2 ) + ( 2 R 2 + x 2 − 2 xR)
(
⎢
2
⎣
dx
2
−7 2
2
−5 ( x − R) ( 2 R 2 + x 2 − 2 xR) ⎤⎥
⎦
Again substituting x =
P30.54
“Helmholtz pair” → separation distance = radius
B=
2 µ0 IR 2
2 ⎡⎣( R 2 ) + R ⎤⎦
2
2
32
=
[
µ0 IR 2
1
4
+ 1]
For N turns in each coil, B =
P30.55
d2B
R
=0 .
and canceling terms,
2
dx 2
32
R
3
=
µ0 I
for 1 turn
1.40 R
−7
µ0 NI ( 4π × 10 ) 100 (10.0 )
=
= 1.80 × 10 −3 T
1.40 R
1.40 ( 0.500 )
Consider first a solid cylindrical rod of radius R carrying current toward you, uniformly distributed over its cross-sectional
area. To find the field at distance r from its center we consider
a circular loop of radius r:
养 B ⋅ d s = µ0 I inside
µ J
µ Jr
k×r
B2π r = µ0 π r 2 J
B= 0
B = 0 k̂
2
2
Now the total field at P inside the saddle coils is the field due
to a solid rod carrying current toward you, centered at the head
of vector a, plus the field of a solid rod centered at the tail of
vector a carrying current away from you.
µ J
µ J
B1 + B2 = 0 kˆ × r1 + 0 −kˆ × r2
2
2
Now note a + r1 = r2
( )
k̂
B1
P
B2
r1
r2
a
kˆ
FIG. P30.55
µ J
µ J
µ Ja
µ J
down in the diagram
k= 0
B1 + B2 = 0 kˆ × r1 − 0 kˆ × ( a + r1 ) = 0 a × k̂
2
2
2
2
192
P30.56
Chapter 30
From Problem 33, the upper sheet creates field
µ J
µ J
B = 0 s kˆ above it and 0 s −k̂ below it.
2
2
y
( )
Consider a patch of the sheet of width w parallel to
the z axis and length d parallel to the x axis. The
d
charge on it σ wd passes a point in time , so
v
w
+
+
+
+
–
x
–
z
–
–
d
FIG. P30.56
σ wv
q σ wd v
and the linear current density is J s =
= σ v. Then
=
w
t
d
1
the magnitude of the magnetic field created by the upper sheet is µ 0σ v. Similarly, the lower
2
sheet in its motion toward the right constitutes current toward the left. It creates magnetic field
1
1
µ 0σ v −k̂ above it and µ 0σ vk̂ below it. We choose to write down answer (b) first.
2
2
the current it constitutes is
( )
(b)
Above both sheets and below both, their equal-magnitude fields add to zero .
(a)
Between the plates, their fields add to
( )
µ0σ v − k̂ = µ0σ v away from you horizontally .
(c)
The upper plate exerts no force on itself. The field of the lower plate,
( )
1
µ 0σ v −k̂
2
will exert a force on the current in the w- by d-section, given by
1
1
I × B = σ wvdˆi × µ 0σ v −kˆ = µ 0σ 2 v 2 wdˆj
2
2
( )
The force per area is
(d)
1 µ0σ 2 v 2 wd
1
µ0σ 2 v 2 up .
ĵ =
2
2
wd
σ
wσ 2 ˆ
− ˆj =
−j .
The electrical force on our section of the upper plate is qE lower = σ w
2∈0
2∈0
( )
The electrical force per area is
we require
v=
( )
wσ 2
σ2
1
σ2
down =
down. To have µ 0σ 2 v 2 =
2 ∈ 0 w
2 ∈0
2
2 ∈0
1
1
=
2
µ0 ∈0
4 π × 10 −7 ( Tm A)( N TAm ) 8.85 × 10 −12 (C2 Nm 2 ) ( As C)
= 3.00 × 10 8 m s
This is the speed of light, not a possible speed for a metal plate.
Sources of the Magnetic Field
193
*P30.57 Model the two wires as straight parallel wires (!). From the treatment
of this situation in the chapter text we have
(a)
FB =
µ0 I 2 2π a
( 4π × 10 ) (140 ) (2π )(0.100 )
=
2π (1.00 × 10 )
−7
FB
2
−3
= 2.46 N upward
P30.58
(b)
The magnetic field at the center of the loop or on
its axis is much weaker than the magnetic field
just outside the wire. The wire has negligible
curvature on the scale of 1 mm, so we model the
lower loop as a long straight wire to find the field
it creates at the location of the upper wire.
(c)
aloop =
(a)
2.46 N − mloop g
mloop
= 107 m s 2 upward
FIG. P30.57
µ
In dB = 0 2 Id s × rˆ , the moving charge constitutes a bit of current as in I = nqv A. For a
4π r
µ
positive charge the direction of d s is the direction of v , so dB = 0 2 nqA ( ds ) v × rˆ . Next,
4π r
A ( ds ) is the volume occupied by the moving charge, and nA ( ds ) = 1 for just one charge. Then,
µ
B = 0 2 qv × rˆ
4π r
( 4 π × 10
−7
T ⋅ m A) (1.60 × 10 −19 C) ( 2.00 × 10 7 m s)
sin 90.0° = 3.20 × 10 −13 T
(b)
B=
(c)
FB = q v × B = (1.60 × 10 −19 C) ( 2.00 × 10 7 m s) ( 3.20 × 10 −13 T) sin 90.0°
4 π (1.00 × 10
)
−3 2
FB = 1.02 × 10 −24 N directed away from the firstt proton
(d)
(8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10−19 C)
kqq
Fe = qE = e 12 2 =
2
r
(1.00 × 10−3 )
2
Fe = 2.30 × 10 −22 N directed away from the firstt proton
Both forces act together. The electrical force is stronger by two orders of magnitude. It is
productive to think about how it would look to an observer in a reference frame moving
along with one proton or the other.
194
P30.59
Chapter 30
−7
µ 0 I ( 4 π × 10 T ⋅ m A) ( 24.0 A)
=
= 2.74 × 10 −4 T
2π r
2 π ( 0.017 5 m )
(a)
B=
(b)
At point C, conductor AB produces a field
produces a field of
( )
1
(2.74 × 10−4 T) −ˆj ,
2
( )
1
(2.74 × 10−4 T) −ˆj ,
2
conductor DE
BD produces no field, and AE produces
( )
negligible field. The total field at C is 2.74 × 10 −4 T − ˆj .
(c)
(d)
(e)
FB = I × B = ( 24.0 A) 0.035 0 mkˆ × ⎡⎣5 ( 2.74 × 10 −4 T) − ˆj ⎤⎦ = (1.15 × 10 −3 N ) ˆi
−3
∑ F 1.15 × 10 N ˆi
= 0.384 m s 2 î
a=
=
m
3.0 × 10 −3 kg
(
(
)
)
( )
(
)
The bar is already so far from AE that it moves through nearly constant magnetic field. The
force acting on the bar is constant, and therefore the bar’s acceleration is constant .
(f)
P30.60
v 2f = vi2 + 2 ax = 0 + 2 ( 0.384 m s 2 ) (1.30 m ) ,
Each turn creates field at the center
so
vf =
(0.999 m s) ˆi
µ0 I
. Together they create field
2R
B
µ 0 I ⎛ 1 1 … 1 ⎞ 4 π × 10 −7 TmI ⎛ 1
1 … 1 ⎞ 1
+ +
+
+ +
⎜ +
⎟=
⎜
⎟
⎝ 5.05 5.15
2 ⎝ R1 R2
R50 ⎠
2A
9.95 ⎠ 10 −2 m
I
FIG. P30.60
= µ 0 I ( 50 m ) 6.93 = 347µ 0 I m
P30.61
µ I
The central wire creates field B = 0 1 counterclockwise. The curved portions of the loop feels no
2π R
force since × B = 0 there. The straight portions both feel I × B forces to the right, amounting
µ I
µ II L
to FB = I 2 2 L 0 1 = 0 1 2 to the right .
πR
2π R
P30.62
(a)
From an equation in the chapter text, the magnetic field produced by one loop at the center
2
µ0 IR 2 µ0 I (π R ) µ0 µ
of the second loop is given by B =
where the magnetic
=
=
2x3
2π x 3
2π x 3
(
)
moment of either loop is µ = I π R 2 . Therefore,
(
2
dB
⎛ µ µ ⎞ ⎛ 3 ⎞ 3µ0 π R I
Fx = µ
= µ⎜ 0 ⎟ ⎜ 4 ⎟ =
⎝ 2π ⎠ ⎝ x ⎠
2π x 4
dx
)
2
=
3π µ0 I 2 R 4
.
2
x4
−7
−3
3π µ 0 I 2 R 4 3π ( 4 π × 10 T ⋅ m A) (10.0 A) ( 5.00 × 10 m )
Fx =
=
= 5.92 × 10 −8 N
4
−2
2
2
x4
(5.00 × 10 m )
2
(b)
4
Sources of the Magnetic Field
P30.63
By symmetry of the arrangement, the magnitude of the net magnetic
field at point P is B = 8 B0 x where B0 is the contribution to the field
L
due to current in an edge length equal to . In order to calculate B0 ,
2
we use the Biot-Savart law and consider the plane of the square to be
the yz-plane with point P on the x-axis. The contribution to the magnetic
field at point P due to a current element of length dz and located a
distance z along the axis is given by the integral form of the Biot-Savart
law as
µ0 I d × rˆ
B0 =
4π ∫ r 2
FIG. P30.63
From the figure we see that
r = x 2 + L2 4 + z 2 and d × rˆ = dz sin θ = dz
(
)
( L 4) + x
L
( 4) + x + z
2
2
2
2
2
By symmetry all components of the field B at P cancel except the components along x
(perpendicular to the plane of the square); and
B0 x = B0 cos φ
Therefore,
cos φ =
where
µ I
B0 x = 0
4π
L 2
∫
0
L2
( L 4) + x
2
sin θ cos φ dz
r2
2
and
B = 8 B0 x
Using the expressions given above for sin θ cos φ , and r, we find
B=
P30.64
µ0 IL2
(
2π x 2 + ( L2 4 )
)
x 2 + ( L2 2 )
There is no contribution from the straight portion of the
wire since d s × rˆ = 0. For the field of the spiral,
µ I ( d s × rˆ )
dB = 0
(4π ) r2
2π
2π
⎡ ⎛ 3π ⎞⎤ 1
d s sin θ rˆ µ 0 I
µ0 I
B=
2 dr ⎢sin ⎜ ⎟⎥ 2
=
∫
∫
2
4 π θ =0
r
4 π θ =0
⎣ ⎝ 4 ⎠⎦ r
(
2π
B=
)
2θ
µ0 I
µ I
∫ r −2 dr = − 40π (r −1 )
4 π θ =0
θ =0
Substitute r = eθ : B = − µ 0 I ⎡e−θ ⎤2 π = − µ 0 I ⎡e−2 π − e0 ⎤ =
⎣ ⎦0
⎣
⎦
4π
4π
µ0 I
(1 − e−2 π ) out of the page.
4π
FIG. P30.64
195
196
P30.65
Chapter 30
Consider the sphere as being built up of little rings of radius r,
centered on the rotation axis. The contribution to the field from
each ring is
dB =
(
µ0 r 2 dI
2 x2 + r2
)
where
32
v
dr
r
dQ ω dQ
dI =
=
2π
t
dx
x
R
dQ = ρ dV = ρ ( 2 π rdr )( dx )
dB =
µ0 ρω r 3 drdx
(
2 x2 + r2
2
+R
R −x
x =− R
r=0
2
∫ ∫
B=
)
where
32
ρ=
Q
( 4 3) π R 3
FIG. P30.65
µ0 ρω r 3 drdx
2 ( x 2 + r 2 )3 2
Let v = r 2 + x 2 , d v = 2 rdr, and r 2 = v − x 2 .
2
R ⎡ R2
R2
⎤
µ 0 ρω ( v − x ) d v
µ 0 ρω
−1 2
2
−3 2
⎥ dx
⎢
B= ∫ ∫
dx
=
d
v
−
x
v
d
v
v
∫
∫
∫
2
2v 3 2
4 x=− R ⎢⎣v=x 2
⎥⎦
x =− R v= x 2
v= x 2
+R
R2
R
⎡
µ 0 ρω
∫ 2v1 2
4 x=− R ⎢⎣
B=
R
⎤
µ 0 ρω ⎡ x 2
2µ 0 ρω
⎢2 − 4 x + 2 R⎥ dx =
∫
4 −R ⎣ R
4
⎦
B=
P30.66
R ⎡
⎛ 1 1 ⎞⎤
⎤
µ 0 ρω
⎢2 ( R − x ) + 2 x 2 ⎜⎜ − ⎟⎟⎥ dx
dx
=
∫
2 ⎥
x ⎦
4 x=− R ⎢⎣
⎝ R x ⎠⎥⎦
B=
R2
x2
+ ( 2 x 2 ) v −1 2
R2
R
⎤
⎡ x2
∫ ⎢⎣2 R − 4 x + 2 R⎥⎦ dx
0
⎞ µ ρω R 2
2µ 0 ρω ⎛ 2 R 3 4 R 2
−
+ 2 R2 ⎟ = 0
⎜
4 ⎝ 3R
2
3
⎠
Consider the sphere as being built up of little rings of radius r,
centered on the rotation axis. The current associated with each
rotating ring of charge is
dI =
v
dr
r
dQ ω ⎡
=
⎣ρ ( 2 π rdr )( dx )⎤⎦
2π
t
dx
x
R
The magnetic moment contributed by this ring is
ω ⎡
3
⎣ρ ( 2 π rdr )( dx )⎤⎦ = πωρ r drdx
2π
dµ = A ( dI ) = π r 2
⎡
⎢
x =− R ⎢
⎣
+R
µ = πωρ
∫
2
R −x
∫
r =0
2
⎤
r 3 dr ⎥ dx = πωρ ∫
⎥⎦
x =− R
+R
(
R −x
2
4
2
)
FIG. P30.66
4
+R
dx = πωρ
∫
x =− R
+R
(R
2
−x
4
µ=
⎛ 3 ⎞ 2 R5 ⎤
πωρ ⎡ 4
πωρ
2 2R
R 4 − 2 R 2 x 2 + x 4 ) dx =
⎢ R ( 2 R) − 2 R ⎜
⎥
⎟+
(
∫
4 x=− R
4 ⎣
⎝ 3 ⎠ 5 ⎦
µ=
πωρ 5 ⎛ 4 2 ⎞ πωρ R 5 ⎛ 16 ⎞ 4 πωρ R 5
R ⎜2 − + ⎟ =
⎜ ⎟=
⎝
4 ⎝ 15 ⎠
15
4
3 5⎠
up
)
2 2
dx
Sources of the Magnetic Field
P30. 67 Note that the current I exists in the conductor with a
I
current density J = , where
A
P1
Bs
r
Bs
2I
Therefore J =
.
π a2
B = Bs − B1 − B2 =
B=
(b)
a/2
µ0 J (π a 2 )
B1 =
a/2
θ
P2
−B1'
θ θ
FIG. P30.67
µ J π (a 2)
µ0 J π (a 2)
.
, and B2 = 0
2 π ( r + ( a 2 ))
2 π ( r − ( a 2 ))
2
⎤
µ J π a2 ⎡1
1
1
⎢ −
⎥
−
2 π ⎢⎣ r 4 ( r − ( a 2 )) 4 ( r + ( a 2 )) ⎦⎥
(
)
⎤
2⎤
⎡ 2
⎥ = µ 0 I ⎢ 2 r − a ⎥ directed to the left
2
⎥⎦
π r ⎣ 4r − a2 ⎦
µ0 J (π a 2 )
2π r
and B1′ = B2′ =
The horizontal components of B1′
B = Bs − B1′ cos θ − B2′ cos θ =
B=
2
r
2
,
2π r
µ 0 ( 2 I ) ⎡⎢ 4 r 2 − a 2 − 2 r 2
2 π ⎢⎣ 4 r r 2 − ( a 2 4 )
At point P2 , Bs =
r + (a 2)
2
To find the field at either point P1 or P2 , find Bs
which would exist if the conductor were solid, using
Ampère’s law. Next, find B1 and B2 that would be
a
due to the conductors of radius
that could occupy
2
the void where the holes exist. Then use the
superposition principle and subtract the field that
would be due to the part of the conductor where the
holes exist from the field of the solid conductor.
At point P1 , Bs =
–B1
–B2
a2 a2 ⎤ π a2
⎡
A = π ⎢a2 − − ⎥ =
4
4⎦
2
⎣
(a)
and
µ0 J (π a 2 )
2π r
µ0 J π (a 2)
2
2π r 2 + (a 2)
2
.
B2′ cancel while their vertical components add.
⎛
⎞
µ0 J π a 2 4 ⎟
⎜
−2
⎜
⎟
2
2
⎝ 2π r + (a 4 ) ⎠
⎤ µ 2I ⎡
2
µ0 J (π a 2 ) ⎡
⎤
r2
⎢1 −
⎥ = 0 ( ) ⎢1 − 2 r ⎥
2
2
2
2
2 π r ⎢⎣ 2 r + ( a 4 ) ⎥⎦
2π r ⎣ 4 r + a ⎦
(
197
)
2⎤
⎡ 2
= µ 0 I ⎢ 2 r 2 + a 2 ⎥ directed towarrd the top of the page
π r ⎣ 4r + a ⎦
r
r + (a2 4 )
2
−B2'
198
Chapter 30
ANSWERS TO EVEN PROBLEMS
P30.2
200 nT
P30.4
1 ⎞ µ0 I
⎛
into the page
⎜⎝ 1 + ⎟⎠
π 2R
P30.6
See the solution.
P30.8
⎛ 1 1 ⎞ µ0 I
into the page
⎜⎝ + ⎟⎠
π 4 2r
P30.10
262 nT into the page
P30.12
(a) 3.84 × 10−21 N up. See the solution. (b) 21.4 mm. This distance is negligible compared to 50
m, so the electron does move in a uniform field. (c) 134 revolutions.
P30.14
(
µ0 I a 2 + d 2 − d a 2 + d 2
2π ad a + d
2
2
) into the page
P30.16
4.97 mm
P30.18
(a) 10.0 µ T out of the page
(b) 80.0 µ N toward wire 1
(c) 16.0 mT into the page
(d) 80.0 µ N toward wire 2
P30.20
(a) It is possible in just one way.
(c) current 2.40 A down
P30.22
(a) 10.0 mm. (b) Yes. If we tried to use wires with diameter 10 mm, they would feel the force
only momentarily after we turn on the current, until they melt. We can use wide, thin sheets of
copper, perhaps plated onto glass, and perhaps with water cooling, to have the forces act
continuously.
P30.24
5.40 cm
P30.26
(a) 400 cm
P30.28
(a) 3.60 T
P30.30
500 A
P30.32
32 × 10 −7 z T ⋅ m
ˆj
(a) See the solution. (b) zero; zero (c) B =
9 × 10 −4 m 2 + z 2
(b) 7.50 nT
(b) wire 3 must be 12.0 cm to the left of wire 1, carrying
(c) 1.26 m
(d) zero
(b) 1.94 T
(d) at d = 3.00 cm (e) 53.3 ĵ mT
P30.34
(a) Make the wire as long and thin as possible without melting when it carries the 5-A current.
Then the solenoid can have many turns. (b) As small in radius as possible with your
experiment fitting inside. Then with a smaller circumference the wire can form a solenoid with
more turns.
P30.36
(a)
P30.38
207 W
P30.40
(a) −Bπ R 2 cos θ
µ0 IN ⎡⎢
a
−
2 ⎢ a2 + R2
⎣
a−
( a − )2
⎤
⎥
+ R 2 ⎥⎦
(b) Bπ R 2 cos θ
(b) See the solution.
Sources of the Magnetic Field
199
P30.42
The field can be uniform in magnitude. Gauss’s law for magnetism implies that magnetic field
lines never start or stop. If the field is uniform in direction, the lines are parallel and their density
stays constant along any one bundle of lines. Thus the magnitude of the field has the same value
at all points along a line in the direction of the field. The magnitude of the field could vary over a
plane perpendicular to the lines, or it could be constant throughout the volume.
P30.44
(a) 12.6 µ T
P30.46
(a) Gravitational field and magnetic field are vectors; atmospheric pressure is a scalar. (b) At
42° north latitude, 76° west longitude, 260 m above sea level: Gravitational field is 9.803 m/s2
down. From the Coast and Geodetic Survey, the magnetic field is 54 mT at 12° west of geographic
north and 69° below the horizontal. Atmospheric pressure is 98 kPa. The atmosphere is held on
by gravitation, but otherwise the effects are all separate. The magnetic field could be produced by
permanent magnetization of a cold iron-nickel deposit within the Earth, so it need not be associated with present-day action of gravitation.
P30.48
2.01 GA west
P30.50
~10 −5 T , enough to affect the compass noticeably
P30.52
µ0 qω
2.5 5π R
P30.54
1.80 mT
P30.56
(a) µ 0σ v horizontally away from you
P30.58
(a) See the solution.
(b) 56.0 µ T
(d) 2.30 × 10 −22
(b) 0
(b) 3.20 × 10 −13 T
(c)
1
µ 0σ 2 v 2 up
2
(d) 3.00 × 10 8 m s
(c) 1.02 × 10 −24 N away from the first proton
N away from the first proton
P30.60
347 µ0 I m perpendicular to the coil
P30.62
(a) See the solution.
P30.64
(m0 I/4p)(1 – e−2p) out of the plane of the paper
P30.66
4
πωρ R 5 upward
15
(b) 59.2 nN
31
Faraday’s Law
CHAPTER OUTLINE
31.1
31.2
31.3
31.4
31.5
31.6
ANSWERS TO QUESTIONS
Faraday’s Law of Induction
Motional emf
Lenz’s Law
Induced emf and Electric Fields
Generators and Motors
Eddy Currents
Q31.1
Magnetic flux measures the “flow” of the magnetic
field through a given area of a loop—even though the
field does not actually flow. By changing the size of the
loop, or the orientation of the loop and the field, one
can change the magnetic flux through the loop, but the
magnetic field will not change.
*Q31.2 The emf is given by the negative of the time derivative
of the magnetic flux. We pick out the steepest downward
slope at instant F as marking the moment of largest emf.
Next comes A. At B and at D the graph line is horizontal so the emf is zero. At E the flux graph slopes gently
upward so the emf is slightly negative. At C the emf has
its greatest negative value. The answer is then F > A >
B = D = 0 > E > C.
*Q31.3 (i) c (ii) b. The magnetic flux is Φ B = BA cos θ . Therefore the flux is a maximum when B is
perpendicular to the loop of wire and zero when there is no component of magnetic field
perpendicular to the loop. The flux is zero when the loop is turned so that the field lies in the
plane of its area.
*Q31.4 (i) Answer (c). The magnetic flux through the coil is constant in time, so the induced emf is zero.
(ii) Answer (a). Positive test charges in the leading and trailing sides of the square experience a
F = q v × B force that is in direction (to the right) × (perpendicularly into the plane away from
you) = toward the top of the square. The charges migrate upward to give positive charge to the
top of the square until there is a downward electric field large enough to prevent more charge
separation.
(
Q31.5
)
By the magnetic force law F = q v × B : the positive charges in the moving bar will feel a
magnetic force in direction (right) × (perpendicularly out of the page) = downward toward the
bottom end of the bar. These charges will move downward and therefore clockwise in the circuit.
If the bar is moving to the left, the positive charge in the bar will flow upward and therefore counterclockwise in the circuit.
(
)
*Q31.6 (i) No. A magnetic force acts within the front and back edges of the coil, but produces no current
and has no influence on the forward motion of the coil.
(ii) Yes. An induced current exists in the bar, which we can attribute either to an induced emf in
the loop or to magnetic force on free charges in the bar. Then a backward magnetic force acts on
the current and an external force must counterbalance it to maintain steady motion.
(iii) No. A magnetic force acts within the bar, but produces no current and has no influence on the
forward motion of the bar.
201
202
Chapter 31
Q31.7
As water falls, it gains speed and kinetic energy. It then pushes against turbine blades, transferring its energy to the rotor coils of a large AC generator. The rotor of the generator turns within a
strong magnetic field. Because the rotor is spinning, the magnetic flux through its turns changes
−NdΦ B
. This
in time as Φ B = BA cos ω t . Generated in the rotor is an induced emf of ε =
dt
induced emf is the voltage driving the current in our electric power lines.
Q31.8
Yes. The induced eddy currents on the surface of the aluminum will slow the descent of the aluminum. In a strong field the plate may fall very slowly.
*Q31.9 Answer (b). The rate of changes of flux of the external magnetic field in the turns of the coil is
doubled, to double the maximum induced emf.
Q31.10 The increasing counterclockwise
current in the solenoid coil produces an upward magnetic field that
increases rapidly. The increasing
upward flux of this field through
the ring induces an emf to produce
clockwise current in the ring. The
magnetic field of the solenoid has a
radially outward component at each
point on the ring. This field component exerts upward force on the
current in the ring there. The whole
ring feels a total upward force larger
than its weight.
FIG. Q31.10
Q31.11
Oscillating current in the solenoid produces an always-changing magnetic field. Vertical flux
through the ring, alternately increasing and decreasing, produces current in it with a direction that
is alternately clockwise and counterclockwise. The current through the ring’s resistance converts
electrically transmitted energy into internal energy at the rate I 2 R.
*Q31.12 (i) Answer (a). The south pole of the magnet produces an upward magnetic field that increases
as the magnet approaches the loop. The loop opposes change by making its own downward
magnetic field; it carries current clockwise, which goes to the left through the resistor.
(ii) Answer (b). The north pole of the magnet produces an upward magnetic field. The loop sees
decreasing upward flux as the magnet falls away, and tries to make an upward magnetic field of
its own by carrying current counterclockwise, to the right in the resistor.
(iii) Answer (c). The north pole of the magnet creates some downward flux through the section
of the coil straight below it, but the south pole creates an equal amount of upward flux. The net
flux through the coil due to the magnet is always zero so the induced emf is zero and the current
is zero.
Faraday’s Law
203
*Q31.13 (i) Answer (b). The battery makes counterclockwise current I1 in the primary coil,
so its magnetic field B1 is to the right and
increasing just after the switch is closed. The
secondary coil will oppose the change
with a leftward field B2, which comes from
an induced clockwise current I 2 that goes
to the right in the resistor. The upper pair of
hands in the diagram represent this effect.
(ii) Answer (d). At steady state the primary
magnetic field is unchanging, so no emf is
induced in the secondary.
(iii) Answer (a). The primary’s field is to
the right and decreasing as the switch
is opened. The secondary coil opposes this
FIG. Q31.13
decrease by making its own field to the right,
carrying counterclockwise current to the left in the resistor. The lower pair of hands diagrammed
represent this chain of events.
*Q31.14 A positive electric charge carried around a circular electric field line in the direction of the field
gains energy from the field every step of the way. It can be a test charge imagined to exist in
vacuum or it can be an actual free charge participating in a current driven by an induced emf. By
doing net work on an object carried around a closed path to its starting point, the magneticallyinduced electric field exerts by definition a nonconservative force. We can get a larger and larger
voltage just by looping a wire around into a coil with more and more turns.
SOLUTIONS TO PROBLEMS
Section 31.1
Faraday’s Law of Induction
Section 31.3
Lenz’s Law
*P31.1
(a)
⎛ B f − Bi ⎞
2 ⎛ 1.5 T − 0 ⎞
⎟1
⎟ cos θ = − ⎡⎣π ( 0.00160 m ) ⎤⎦ ⎜
⎝ 0.120 s ⎠
⎝ ∆t ⎠
ε = − N ∆BA cos θ = −1π r 2 ⎜
∆t
= −(8.04 × 10 −6 m 2 )12.5 T/s
= 101 mV tending to produce clockwise current as seen from above
(b)
P31.2
The rate of change of magnetic field in this case is (–0.5 T – 1.5 T)/0.08 s = 25 T/s. It is
twice as large in magnitude and in the opposite sense from the rate of change in case (a), so
the emf is also twice as large in magnitude and in the opposite sense .
ε
−4
2
∆Φ B ∆ B ⋅ A ( 2.50 T − 0.500 T) (8.00 × 10 m ) ⎛ 1 N ⋅ s ⎞ ⎛ 1 V ⋅ C ⎞
=
=
=
⎜
⎟⎜
⎟
⎝1 T ⋅ C ⋅ m ⎠ ⎝1 N ⋅ m ⎠
∆t
∆t
1.00 s
ε
= 1.60 mV
(
)
and
I loop =
ε = 1.60 mV =
R
2.00 Ω
0.800 mA
204
P31.3
Chapter 31
⎛ cos θ f − cos θ i ⎞
⎟
∆t
⎝
⎠
ε = − N ∆BA cos θ = − NBπ r 2 ⎜
∆t
2 ⎛ cos 180° − cos 0° ⎞
= −25..0 (50.0 × 10 −6 T) ⎡⎣π ( 0.500 m ) ⎤⎦ ⎜
⎟
⎝
⎠
0.200 s
*P31.4
ε=
+9.82 mV
(a)
We evaluate the average emf:
⎛ B f − Bi ⎞
2 ⎛ 0 − 0.11 T ⎞
⎟ cos 0°
⎟ cos θ = −12 ⎡⎣π ( 0.0210 m ) ⎤⎦ ⎜
⎝ 0.180 s ⎠
⎝ ∆t ⎠
ε = − N ∆BA cos θ = − N π r 2 ⎜
∆t
= 0.00102 V
The average induced current will then be 0.0102 V/2.3 Ω = 4.42 mA. If the meter has a
sufficiently short response time, it will register the current. The average current may even
run the meter offscale by a factor of 4.42, so you might wish to slow down the motion of
the coil.
(b)
P31.5
*P31.6
Positive. The coil sees decreasing external magnetic flux toward you, so it makes some flux
of its own in this direction by carrying counterclockwise current, that enters the red terminal of the ammeter.
(a)
ε = − dΦ B = − A dB =
(b)
ε=(
(c)
At t = 0
dt
dt
ABmax −t τ
e
τ
0.160 m 2 ) ( 0.350 T)
2.00 s
ε=
ε = − N dΦ B = − N
dt
e−4.00 2.00 = 3.79 mV
28.0 mV
d
( BA cosθ )
dt
ε = − NB cosθ ⎛⎜ ∆A ⎞⎟ = −200 (50.0 × 10 −6 T ) ( cos 62.0°) ⎛⎜ 39.0 × 10
⎝
⎝ ∆t ⎠
P31.7
−4
m2 ⎞
⎟⎠ = −10.2 µ V
1.80 s
Noting unit conversions from F = qv × B and U = qV , the induced voltage is
ε = −N
d B⋅A
(
dt
= 3 200 V
I=
) = − N ⎛ 0 − B A cosθ ⎞ = +200 (1.60 T)( 0.200 m ) cos 0° ⎛
ε = 3 200 V =
R
20.0 Ω
2
⎝
160 A
i
∆t
⎠
20.0 × 10
−3
s
1 N ⋅ s ⎞ ⎛ 1 V ⋅ C⎞
⎝1 T⋅C⋅m⎠ ⎝ N ⋅m ⎠
Faraday’s Law
P31.8
205
h+ w
µ I
µ IL dx
µ IL ⎛ h + w ⎞
d Φ B = B ⋅ dA = 0 Ldx : Φ B = ∫ 0
= 0 ln ⎜
⎟
⎝ h ⎠
2π x
2π x
2π
h
(a)
⎤
⎡
⎤
⎡
ε = − dΦ B = − d ⎢ µ0 IL ln ⎛⎜ h + w ⎞⎟⎥ = −⎢ µ0 L ln ⎛⎜ h + w ⎞⎟⎥ dI
(b)
dt ⎣ 2 π
dt
4 π × 10
ε =−(
−7
⎝ h ⎠⎦
T ⋅ m A) (1.00 m )
2π
⎣ 2π
⎝ h ⎠⎦ dt
⎛ 1.00 + 10.0 ⎞
ln ⎜
⎟ (10.0 A s)
⎝ 1.00 ⎠
= −4.80 µV
The long wire produces magnetic flux into the page through the rectangle,
shown by the first hand in the figure to the right.
As the magnetic flux increases, the rectangle produces its own magnetic
field out of the page, which it does by carrying counterclockwise
current (second hand in the figure).
P31.9
ε
=
d ( BA)
dI
= 0.500µ 0 nA = 0.480 × 10 −3 V
dt
dt
I ring =
(a)
ε = 4.80 × 10−4 =
R
3.00 × 10 −4
1.60 A counterclockwise
I
µ0 I
= 20.1 µ T
2rring
5 cm
(b)
Bring =
(c)
The coil’s field points into the coil, and is increasing,
so
P31.10
ε
(a)
=
FIG. P31.8
Bring points away from the center of the coiil, or left
in the textbook picture
Iring
3 cm
FIG. P31.9
.
d ( BA)
dI
∆I
= 0.500µ 0 nA = 0.500µ 0 nπ r22
dt
dt
∆t
I ring =
ε=
R
µ 0 nπ r22 ∆I
2 R ∆t
counterclockwise
I
r1
Iring
r2
µ0 I
µ nπ r ∆I
=
2 r1
4 r1 R ∆t
2
0
2
2
(b)
B=
(c)
The coil’s field points into the coil, and is increasing,
FIG. P31.10
so
Bring points away from the center of the coiil, or left
in the textbook picture
.
206
P31.11
Chapter 31
Φ B = ( µ0 nI ) Asolenoid
2
ε = − N dΦ B = − N µ0 n (π rsolenoid
) dI
dt
dt
ε = −15.0 ( 4 π × 10 T ⋅ m A) (1.00 × 10 3 m −1 ) π (0.020 0 m )2 (600
ε = −14.2 cos (120t ) mV
−7
ε = −20(130)4π × 10−7 (0 − 6[1]/13 × 10−6 )π (0.03)2 1/00.8 =
*P31.12 (a)
(b)
P31.13
A s) cos (120t )
5.33 V
A flat compact circular coil with 130 turns and radius 40.0 cm carries current 3.00 A counterclockwise. The current is smoothly reversed to become 3.00 A clockwise after 13.0 ms.
At the center of this primary coil is a secondary coil in the same plane, with 20 turns and
radius 3.00 cm. Find the emf induced in the secondary.
For a counterclockwise trip around the left-hand loop,
with B = At
d
⎡ At ( 2a 2 ) cos 0° ⎤⎦ − I1 ( 5 R ) − I PQ R = 0
dt ⎣
and for the right-hand loop,
d
⎡ Ata 2 ⎤⎦ + I PQ R − I 2 ( 3 R ) = 0
dt ⎣
where I PQ = I1 − I 2 is the upward current in QP.
)
(
Thus,
2 Aa 2 − 5 R I PQ + I 2 − I PQ R = 0
and
Aa 2 + I PQ R = I 2 ( 3 R )
2 Aa 2 − 6 RI PQ −
I PQ =
I PQ
P31.14
ε
=
(
)
5
Aa 2 + I PQ R = 0
3
Aa 2
upward, and since R = (0.100 Ω/m)(0.650 m) = 0.0650 Ω
23 R
(1.00 × 10
=
−3
T s) ( 0.650 m )
23 ( 0.065 0 Ω)
2
= 283 µ A upward .
⎛ dB ⎞
∆Φ B
= N ⎜ ⎟ A = N ( 0.010 0 + 0.080 0t ) A
⎝ dt ⎠
∆t
At t = 5.00 s,
P31.15
FIG. P31.13
ε
2
= 30.0 ( 0.410 T s) ⎡⎣π ( 0.040 0 m ) ⎤⎦ = 61.8 mV
B = µ0 nI = µ0 n ( 30.0 A ) (1 − e−1.60 t )
Φ B = ∫ BdA = µ0 n ( 30.0 A ) (1 − e−1.60 t ) ∫ dA
Φ B = µ0 n ( 30.0 A ) (1 − e−1.60 t ) π R 2
ε = − N dΦ B = − N µ0 n ( 30.0 A) π R2 (1.60) e−1.60t
dt
ε = − (250) ( 4 π × 10−7
ε = (68.2 mV) e−1.60t
FIG. P31.15
2
N A 2 )( 400 m −1 ) ( 30.0 A) ⎡⎣π ( 0.060 0 m ) ⎤⎦1.60 s−1e−1.60 t
counterclockwise
Faraday’s Law
P31.16
(a)
207
Suppose, first, that the central wire is long and straight. The enclosed current of unknown
amplitude creates a circular magnetic field around it, with the magnitude of the field given
by Ampère’s law.
µ0 I max sin ω t
2π R
at the location of the Rogowski coil, which we assume is centered on the wire. This field
passes perpendicularly through each turn of the toroid, producing flux
∫ B ⋅ d s = µ I :
B=
0
µ I A sin ω t
B ⋅ A = 0 max
2π R
The toroid has 2π Rn turns. As the magnetic field varies, the emf induced in it is
ε = −N
µ I A d
d B ⋅ A = −2π Rn 0 max
sin ω t = − µ0 I max nAω cos ω t
dt
2π R dt
This is an alternating voltage with amplitude ε max = µ 0 nAω I max . Measuring the amplitude
determines the size I max of the central current. Our assumptions that the central wire is
long and straight and passes perpendicularly through the center of the Rogowski coil are all
unnecessary.
(b)
If the wire is not centered, the coil will respond to stronger magnetic fields on one side,
but to correspondingly weaker fields on the opposite side. The emf induced in the coil is
proportional to the line integral of the magnetic field around the circular axis of the toroid.
Ampère’s law says that this line integral depends only on the amount of current the coil
encloses. It does not depend on the shape or location of the current within the coil, or on
any currents outside the coil.
ε = d ( NB2 cosθ ) = N ∆B cosθ
dt
∆t
2
P31.17
ε
∆t
=
N ∆B cos θ
=
(80.0 × 10
(50) (600 × 10
−6
−3
V) ( 0.400 s)
T − 200 × 10 −6 T) cos ( 30.0°)
= 1.36 m
Length = 4 N = 4 (1.36 m ) ( 50 ) = 272 m
P31.18
In a toroid, all the flux is confined to the inside of the
toroid.
B=
µ 0 NI 500µ 0 I
=
2π r
2π r
Φ B = ∫ BdA =
ΦB =
500 µ0 I max
adr
sin ω t ∫
2π
r
500 µ0 I max
b + R⎞
a sin ω t ln ⎛
⎝
2π
R ⎠
ε = N 2 dΦ B = 20 ⎛⎜ 500µ0 I max ⎞⎟ω a ln ⎛⎜ b + R ⎞⎟ cos ω t
dt
⎝
4
ε = 10 ( 4 π × 10−7 N
2π
=
(0.422 V) cos ω t
2π
⎠
⎝ R ⎠
FIG. P31.18
⎛ ( 3.00 + 4.00 ) cm ⎞
A 2 ) ( 50.0 A)( 377 rad s) ( 0.020 0 m ) ln ⎜
⎟ cos ω t
4.00 cm
⎠
⎝
208
P31.19
Chapter 31
The upper loop has area π ( 0.05 m ) = 7.85 × 10 −3 m 2 . The induced emf in it is
2
ε = −N
d
dB
BA cos θ = −1A cos 0°
= −7.85 × 10 −3 m 2 ( 2 T s ) = −1.57 × 10 −2 V
dt
dt
The minus sign indicates that it tends to produce counterclockwise current, to make its own
magnetic field out of the page. Similarly, the induced emf in the lower loop is
ε = − NA cosθ dB = −π (0.09 m )2 2
dt
T s = −5.09 × 10 −2 V = +5.09 × 10 −2 V to produce
counterclockwise current in the lower loop, which becomes clockwise current in the uppper loop .
The net emf for current in this sense around the figure 8 is
5.09 × 10 −2 V − 1.57 × 10 −2 V = 3.52 × 10 −2 V.
It pushes current in this sense through series resistance [ 2π ( 0.05 m ) + 2π ( 0.09 m )] 3 Ω m = 2.64 Ω.
The current is I =
ε = 3.52 × 10−2 V =
2.64 Ω
R
Section 31.2
Motional emf
Section 31.3
Lenz’s Law
P31.20
(a)
13.3 mA .
For maximum induced emf, with positive charge at the top of the antenna,
F+ = q+ v × B , so the auto must move east .
(
)
⎛
m⎞
−4
⎟ cos 65.0° = 4.58 × 10 V
3 600 s ⎠
ε = Bv = (5.00 × 10−5 T) (1.20 m ) ⎜ 65.0 × 10
(b)
⎝
3
ε = Bv = (1.2 × 10−6 T) (14.0 m )(70 m/s) =
1.18 × 10 −3 V . A free positive test charge
in the wing feels a magnetic force in direction v × B = north × down = west. Then it
migrates west to make the left-hand wingtip positive.
*P31.21 (a)
(b) No change. The charges in the horizontally-moving wing respond only to the vertical
component of the Earth’s field.
(c) No. If we tried to connect the wings into a circuit with the light bulb, we would run an extra
insulated wire along the wing. With the wing it would form a one-turn coil, in which the
emf is zero as the coil moves in a uniform field.
P31.22
I=
ε = Bv = 2.5(1.2)v/ 6 = 0.5
R
R
v = 1.00 m s
FIG. P31.22
Faraday’s Law
P31.23
209
FB = I × B = I B
(a)
ε
When
I=
and
ε = Bv
we get
FB =
R
Bv
B 2 2 v ( 2.50 ) (1.20 ) ( 2.00 )
=
( B) =
6.00
R
R
= 3.00 N.
2
2
FIG. P31.23
The applied force is 3.00 N to the right .
P = I 2R =
(b)
*P31.24 FB = I B
and
ε = Bv
so
B=
or
P = Fv = 6.00 W
I=
ε = Bv
(a)
FB =
(b)
I 2 R = 2.00 W
(c)
For constant force, P = F ⋅ v = (1.00 N ) ( 2.00 m s ) = 2.00 W .
(d)
P31.25
B2 2 v2
= 6.00 W
R
R
R
I 2 R
v
and
IR
v
I=
FB v
= 0.500 A
R
The powers computed in parts (b) and (c) are mathematically equal. More profoundly, they
are physically identical. Each bit of energy delivered to the circuit mechanically immediately
goes through being electrically transmitted to the resistor and there becomes additional internal
energy. Counting the 2 W twice is like counting your lunch money twice.
Observe that the homopolar generator has no commutator and
produces a voltage constant in time: DC with no ripple. In time
dt, the disk turns by angle dθ = ω dt . The outer brush slides over
distance rdθ . The radial line to the outer brush sweeps over area
1
1
dA = rrdθ = r 2ω dt
2
2
d The emf generated is
ε = −N B ⋅ A
dt
ε = − (1) B cos 0 º dA = − B ⎛⎝ 1 r 2ω ⎞⎠
dt
2
(We could think of this as following from the result of the
example in the chapter text about the helicopter blade.)
The magnitude of the emf is
ε
⎤ ⎛ 2 π rad rev ⎞
⎛1
⎞
⎡1
2
= B ⎜ r 2ω ⎟ = ( 0.9 N ⋅ s C ⋅ m ) ⎢ ( 0.4 m ) (3 200 reev min )⎥ ⎜
⎟
⎦ ⎝ 60 s min ⎠
⎣2
⎝2
⎠
ε
= 24.1 V
FIG. P31.25
A free positive charge q shown, turning with the disk, feels a magnetic force qv × B
radially outward. Thus the outer contact is positive .
210
P31.26
Chapter 31
The speed of waves on the wire is
T
267 N ⋅ m
=
= 298 m s
µ
3 × 10 −3 kg
v=
In the simplest standing-wave vibration state,
λ
d NN = 0.64 m =
λ = 1.28 m
2
v 298 m s
and
f= =
= 233 Hz
λ 1.28 m
(a) The changing flux of magnetic field through the circuit containing the wire will drive
current to the left in the wire as it moves up and to the right as it moves down. The emf
will have this same frequency of 233 Hz .
(b)
The vertical coordinate of the center of the wire is described by
x = A cos ω t = (1.5 cm ) cos ( 2π 233 t s ) .
dx
= − (1.5 cm )( 2 π 233 s) sin ( 2 π 233 t s) .
dt
Its maximum speed is 1.5 cm ( 2π ) 233 s = 22.0 m s .
Its velocity is v =
The induced emf is ε = −Bv, with amplitude
εmax = Bvmax = 4.50 × 10−3 T (0.02 m ) 22 m s = 1.98 × 10−3 V .
P31.27
ω = ( 2.00 rev s )( 2π rad rev) = 4.00π rad s
ε = 1 Bω 2 =
2
P31.28
(a)
(b)
2.83 mV
Bext = Bext ˆi and Bext decreases; therefore,
the induced field is B0 = B0 ˆi (to the right)
and the current in the resistor is directed
to the right .
Bext = Bext − ˆi increases; therefore, the
induced field B = B + ˆi is to the right,
( )
0
(c)
(d)
0
( )
and the current in the resistor is directed
out of the plane in the textbook picture and
to the right in the diagram here.
Bext = Bext − kˆ into the paper and Bext
decreases; therefore, the induced field is
B0 = B0 − kˆ into the paper, and the current in
( )
( )
the resistor is directed to the right .
By the magnetic force law, FB = q v × B .
Therefore, a positive charge will move to the
top of the bar if B is into the paper .
(
)
FIG. P31.28
Faraday’s Law
P31.29
(a)
211
The force on the side of the coil entering the field
(consisting of N wires) is
F = N ( ILB) = N ( IwB)
The induced emf in the coil is
ε
=N
dΦ B
d ( Bwx )
=N
= NBwv
dt
dt
so the current is I =
ε
R
=
NBwv
counterclockwise.
R
The force on the leading side of the coil is then:
F = N⎛
⎝
(b)
NBwv ⎞
N 2 B2 w2 v
wB =
to the left
R ⎠
R
Once the coil is entirely inside the field,
Φ B = NBA = constant,
so
(c)
ε = 0,
I = 0,
and
F= 0
FIG. P31.29
As the coil starts to leave the field, the flux decreases at the rate Bwv, so the magnitude
of the current is the same as in part (a), but now the current is clockwise. Thus, the force
exerted on the trailing side of the coil is:
F=
N 2 B2 w2 v
to the left again
R
P31.30
Look in the direction of ba. The bar magnet creates a field into the page, and the field increases.
The loop will create a field out of the page by carrying a counterclockwise current. Therefore,
current must flow from b to a through the resistor. Hence, Va − Vb will be negative .
P31.31
Name the currents as shown in the diagram:
Left loop:
+ Bd v2 − I 2 R2 − I1 R1 = 0
Right loop:
+ Bd v3 − I 3 R3 + I1 R1 = 0
At the junction:
I 2 = I1 + I 3
Then,
Bd v2 − I1 R2 − I 3 R2 − I1 R1 = 0
I3 =
So,
Bd v3 I1 R1
+
R3
R3
FIG. P31.31
Bd v2 − I1 ( R1 + R2 ) −
Bd v3 R2 I1 R1 R2
−
=0
R3
R3
⎛
⎞
v2 R3 − v3 R2
I1 = Bd ⎜
⎟ upward
⎝ R1 R2 + R1 R3 + R2 R3 ⎠
⎡
⎤
( 4.00 m s)(15.0 Ω) − (2.00 m s)(10.0 Ω)
I1 = ( 0.010 0 T)( 0.100 m ) ⎢
⎥ = 145 µ A
⎢⎣ ( 5.00 Ω)(10.0 Ω) + ( 5.00 Ω) (15.0 Ω) + (10.0 Ω)(15.0 Ω) ⎦⎥
upward.
212
Chapter 31
Section 31.4
P31.32
(a)
Induced emf and Electric Fields
dB
= 6.00t 2 − 8.00t
dt
ε
At t = 2.00 s,
E=
dΦ B
dt
π R 2 ( dB dt )
=
=
2 π r2
8.00 π ( 0.0250 )
2
2 π ( 0.0500 )
F = qE = 8.00 × 10 −21 N
(b)
P31.33
When 6.00t 2 − 8.00t = 0,
ε
dB
= 0.060 0t
dt
=
FIG. P31.32
t = 1.33 s
dΦ B
dB
= π r12
= 2 π r1 E
dt
dt
At t = 3.00 s,
⎛ π r 2 ⎞ dB 0.020 0 m
⎛ 1 N⋅s ⎞
0.060 0 T s 2 ) ( 3.00 s) ⎜
E =⎜ 1 ⎟
=
⎟
(
⎝1 T ⋅ C ⋅ m ⎠
2
⎝ 2 π r1 ⎠ dt
E = 1.80 × 10 −3 N C perpendicular to r1 and counterclockwise
P31.34
(a)
(b)
Section 31.5
P31.35
FIG. P31.33
dΦ B
dt
2 dB
2 π rE = ( π r )
dt
∫ E ⋅ d =
so
E=
(9.87
mV m ) cos (100π t )
The E field is always opposite to increasing B. Therefore it is clockwise .
Generators and Motors
(a)
ε max = NABω = (1 000 ) ( 0.100 ) ( 0.200 ) (120π ) =
(b)
ε (t ) = NBAω sin ω t = NBAω sin θ
ε
7.54 kV
V
is maximal when sin θ = 1
or θ = ±
π
2
FIG. P31.35
so the plane of coil is parallel to B .
P31.36
⎛ 2π rad ⎞ ⎛ 1 min ⎞
= 314 rad s
For the alternator, ω = ( 3 000 rev min ) ⎜
⎝ 1 rev ⎟⎠ ⎝⎜ 60 s ⎟⎠
ε = − N d Φ B = −250 d ⎡⎣( 2.50 × 10 −4 T ⋅ m 2 ) cos (314t s )⎤⎦ = +250 ( 2.50 × 10 −4 T ⋅ m 2 ) (314 s ) sin (314t )
dt
dt
(a)
ε = (19.6 V ) sin (314t )
(b)
ε max = 19.6 V
Faraday’s Law
P31.37
213
B = µ 0 nI = ( 4 π × 10 −7 T ⋅ m A)( 200 m −1 ) (15.0 A) = 3.77 × 10 −3 T
For the small coil, Φ B = NB ⋅ A = NBA cos ω t = NB ( π r 2 ) cos ω t .
ε = − dΦ B = NBπ r 2ω sin ω t
Thus,
dt
ε = (30.0 ) (3.77 × 10 −3 T ) π ( 0.080 0 m )2 ( 4.00π s−1 ) sin ( 4.00π t ) =
P31.38
As the magnet rotates, the flux
through the coil varies sinusoidally
in time with Φ B = 0 at t = 0.
Choosing the flux as positive
when the field passes from left
to right through the area of the
coil, the flux at any time may be
written as Φ B = − Φ max sin ω t
so the induced emf is given by
ε = − d Φ B = ω Φ max cos ω t .
dt
The current in the coil is then
ε ω Φ max cos ω t = I cos ω t .
I= =
max
R
R
( 28.6 mV ) sin ( 4.00π t )
1
0.5
I/Imax
0
0
0.5
1
1.5
–0.5
–1
t/T = (wt/2p)
FIG. P31.38
850 mA
P31.39
120 V
M
To analyze the actual circuit, we model it as
850 mA
120 V
(a)
11.8 Ω
e back
The loop rule gives +120 V − 0.85 A (11.8 Ω ) − ε back = 0
ε back =
110 V
(b)
The resistor is the device changing electrical work input into internal energy:
P = I 2 R = ( 0.85 A )2 (11.8 Ω ) = 8.53 W
(c)
With no motion, the motor does not function as a generator, and
120 V − I c (11.8 Ω) = 0
I c = 10.2 A
Pc = I c2 R = (10.2 A) (11.8 Ω) = 1.22 kW
2
ε back = 0. Then
2
214
P31.40
Chapter 31
(a)
Φ B = BA cos θ = BA cos ω t = ( 0.800 T ) ( 0.010 0 m 2 ) cos 2π ( 60.0 ) t
=
2
(b)
ε = − dΦ B =
(c)
I=
(d)
P = I 2 R = ( 9.10 W ) sin 2 ( 377t )
(e)
P = Fv = τω
dt
Section 31.6
P31.41
(8.00 mT ⋅ m ) cos (377t )
ε=
R
( 3.02 V ) sin ( 377t )
( 3.02 A ) sin ( 377t )
τ=
so
P
ω
=
(24.1 mN ⋅ m ) sin 2 ( 377t )
Eddy Currents
The current in the magnet creates an
upward magnetic field, so the N and S poles on
the solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B
increases as the coil approaches, so a current is induced here to create a downward magnetic field.
This is
clockwise current, so the S pole on the rail is shown correctly. On the rail behind
the brake, the upward magnetic flux is decreasing. The induced current in the rail will produce
upward magnetic field by being
P31.42
(a)
counterclockwise as the picture correctly shows.
Consider an annulus of radius r, width dr, height b, and resistivity r. Around its circumference, a voltage is induced according to
d d
B ⋅ A = −1 Bmax ( cos ω t ) π r 2 = + Bmaxπ r 2ω siin ω t
dt
dt
ρ ρ ( 2π r )
=
The resistance around the loop is
Ax
bdr
ε = −N
The eddy current in the ring is dI =
The instantaneous power is
ε
resistance
=
Bmaxπ r 2ω ( sin ω t ) bdr Bmax rbω dr sin ω t
=
2ρ
ρ ( 2π r )
2
π r 3bω 2 dr sin 2 ω t
Bmax
2ρ
1
1
1
1
sin 2 ω t = − cos 2ω t is − 0 =
2 2
2
2
dPi = ε dI =
The time average of the function
so the time-averaged power delivered to the annulus is
dP =
2
π r 3bω 2 dr
Bmax
4ρ
The power delivered to the disk is
R
P = ∫ dP = ∫
0
P=
π B R bω
B π bω ⎛ R 4
⎞
⎜⎝ 4 − 0⎟⎠ =
4ρ
16 ρ
2
max
2
2
max
4
2
π bω 2 3
Bmax
r dr
4ρ
2
(b)
When Bmax gets two times larger,
2
Bmax
and P get 4 times larger.
(c)
When f and ω = 2 π f double,
ω 2 and P get 4 times larger.
(d)
When R doubles,
R 4 and P become 2 4 = 16 times larger.
Faraday’s Law
P31.43
(a)
At terminal speed,
⎛ε ⎞
⎛ BwvT
Mg = FB = IwB = ⎜ ⎟ wB = ⎜
⎝R⎠
⎝ R
vT =
or
(b)
215
⎞
B 2 w 2 vT
⎟ wB =
⎠
R
MgR
B 2ω 2
The emf is directly proportional to vT , but the
current is inversely proportional to R. A large R
means a small current at a given speed, so the loop
must travel faster to get FB = mg.
FIG. P31.43
(c)
At a given speed, the current is directly proportional to the magnetic field. But the force
is proportional to the product of the current and the field. For a small B, the speed must
increase to compensate for both the small B and also the current, so vT ∝ 1/ B 2.
Additional Problems
*P31.44 (a)
The circuit encloses increasing flux of magnetic field into the page, so it tries to make its
own field out of the page, by carrying counterclockwise current.
(b)
I = e /R = Bv/R = (0.4 T 0.8 m 15 m/s)/48 Ω = 100 mA
(c)
The magnetic field exerts a backward magnetic force on the induced current. With the
values in (b), this force is IB = 0.1 A 0.8 m 0.4 T = 0.032 N, much less than 0.6 N. The
speed of the bar increases until the backward magnetic force exerted on the current in the
bar is equal to 0.6 N. The terminal speed is given by 0.6 N = IB = (B)2v/R.
Then v = 0.6 N R/(B)2 = 0.6 N 48 Ω/(0.4 T 0.8 m)2 = 281 m /s .
P31.45
(d)
Fv = 0.6 N 281 m/s = 169 W
(e)
The terminal speed becomes larger . The bar must move faster to generate a larger emf to
produce enough current in the larger resistance to feel the 0.6-N magnetic force.
(f)
The power delivered to the circuit by the agent moving the bar, and then converted into
internal energy by the resistor, is described by P = F v = F 2 R / B 2 2. Thus the power is
directly proportional to the resistance and becomes larger as the bulb heats up.
ε = − N d ( BA cos θ ) = − N (π r 2 ) cos 0° ⎛⎜⎝ dB ⎞⎟⎠
dt
dt
ε = − ( 30.0) ⎡⎣⎢π (2.70 × 10−3 m )2 ⎤⎦⎥(1) d ⎡⎣50.0 mT + ( 3.20 mT) sin (2π ⎡⎣523t s−1 ⎤⎦ )⎤⎦
dt
2⎤
⎡
−3
ε = − ( 30.0) ⎢⎣π (2.70 × 10 m ) ⎥⎦( 3.20 × 10−3 T) ⎡⎢⎣2π (523 s−1 ) cos 2π ⎡⎣523t s−1 ⎤⎦ ⎤⎦
(
ε = − ( 7.22 × 10−3 V) cos ⎡⎣2π (523t s−1 )⎤⎦
)
216
P31.46
P31.47
Chapter 31
ε = −N
⎛ 1.50 T − 5.00 T ⎞
∆
∆B
( BA cos θ ) = − N ( π r 2 ) cos° = −1 ( 0.005 00 m 2 ) (1) ⎜
⎟ = 0.875 V
−3
∆t
∆t
⎝ 20.0 × 10 s ⎠
ε=
0.875 V
= 43.8 A
0.020 0 Ω
(a)
I=
(b)
P = ε I = ( 0.875 V ) ( 43.8 A ) = 38.3 W
(a)
Doubling the number of turns has this effect:
R
Amplitude doubles and period unchanged
(b)
Doubling the angular velocity has this effect:
doubles the amplitude and cuts the period inn half
(c)
Doubling the angular velocity while reducing
the number of turns to one half the original value
has this effect:
FIG. P31.47
Amplitude unchanged and period is cut in hallf
P31.48
In the loop on the left, the induced emf is
ε = dΦ B = A dB = π (0.100 m )2 (100 T s) = π V
dt
dt
and it attempts to produce a counterclockwise current
in this loop.
In the loop on the right, the induced emf is
ε
=
dΦ B
2
= π ( 0.150 m ) (100 T s) = 2.25π V
dt
FIG. P31.48
and it attempts to produce a clockwise current. Assume that I1 flows down through the 6.00-Ω
resistor, I 2 flows down through the 5.00-Ω resistor, and that I 3 flows up through the 3.00-Ω
resistor.
From Kirchhoff’s junction rule:
I 3 = I1 + I 2
(1)
Using the loop rule on the left loop:
6.00 I1 + 3.00 I 3 = π
(2)
Using the loop rule on the right loop:
5.00 I 2 + 3.00 I 3 = 2.25π
(3)
Solving these three equations simultaneously, I3 = (p – 3I3)/6 + (2.25 p – 3I3)/5
I1 = 0.062 3 A , I 2 = 0.860 A , and I 3 = 0.923 A .
Faraday’s Law
P31.49
The emf induced between the ends of the moving bar is
ε = Bv = ( 2.50 T ) ( 0.350 m ) (8.00
m s ) = 7.00 V
The left-hand loop contains decreasing flux away from you, so the induced current in it will be
clockwise, to produce its own field directed away from you. Let I1 represent the current
flowing upward through the 2.00-Ω resistor. The right-hand loop will carry counterclockwise
current. Let I 3 be the upward current in the 5.00-Ω resistor.
(a)
Kirchhoff’s loop rule then gives: +7.00 V − I1 ( 2.00 Ω ) = 0
I1 = 3.50 A
+7.00 V − I 3 ( 5.00 Ω ) = 0
I 3 = 1.40 A
and
(b)
The total power converted in the resistors of the circuit is
P = ε I1 + ε I 3 = ε ( I1 + I 3 ) = ( 7.00 V ) ( 3.50 A + 1.40 A ) = 34.3 W
(c)
Method 1: The current in the sliding conductor is downward with value
I 2 = 3.50 A + 1.40 A = 4.90 A. The magnetic field exerts a force of
Fm = I B = ( 4.90 A )( 0.350 m )( 2.50 T) = 4.29 N directed
toward the right
on this conductor. An outside agent must then exert a force of 4.29 N to the left to keep
the bar moving.
Method
2: The agent moving the bar must supply the power according to
P = F ⋅ v = Fv cos 0°. The force required is then:
F=
P31.50
I=
P
v
=
ε + εinduced
R
34.3 W
= 4.29 N
8.00 m s
and
εinduced = − d ( BA)
dt
dv
F=m
= IBd
dt
d v IBd Bd
=
=
(ε + εinduced )
dt
m
mR
d v Bd
=
(ε − Bvd )
dt mR
To solve the differential equation, let u = ε − Bvd
du
dv
= − Bd
dt
dt
1 du Bd
−
=
u
Bd dt mR
du
( Bd )
∫u u = − ∫0 mR dt
0
u
so
t
2
u
( Bd ) t
=−
u0
mR
2
Integrating from t = 0 to t = t ,
or
Since v = 0 when t = 0,
and
ln
2 2
u
= e− B d t mR
u0
u0 = ε
u = ε − Bv d
ε − Bvd = ε e− B d t mR
2 2
Therefore,
v=
ε
Bd
(1 − e
− B 2 d 2t mR
)
FIG. P31.50
217
218
P31.51
Chapter 31
Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose
we use a bar magnet to produce field 10 −3 T through the coil in one direction along its axis.
Suppose we then flip the magnet to reverse the flux in 10 −1 s. The average induced emf is then
ε = − N ∆Φ B = − N ∆ [ BA cosθ ] = − NB (π r 2 ) ⎛⎝ cos180° − cos 0° ⎞⎠
∆t
∆t
∆t
ε = − ( 20 ) (10 −3 T ) π ( 0.0150 m )2 ⎛⎝
P31.52
I=
(a)
dq ε
=
where
dt R
−2 ⎞
~ 10 −4 V
10 −1 s ⎠
ε = −N dΦ B
dt
∫ dq =
so
N
R
Φ2
∫ dΦ
B
Φ1
and the charge passing any point in the circuit will be Q =
Q=
(b)
N
R
so B =
P31.53
I=
⎡
⎛ π ⎞ ⎤ BAN
⎢ BA cos 0 − BA cos ⎜⎝ 2 ⎟⎠ ⎥ = R
⎦
⎣
−4
RQ ( 200 Ω ) ( 5.00 × 10 C )
=
= 0.250 T
NA (100 ) ( 40.0 × 10 −4 m 2 )
ε=B
R
A
R ∆t
so q = I ∆t =
P31.54
N
( Φ 2 − Φ1 ) .
R
(15.0 µ T) (0.200 m )2
0.500 Ω
= 1.20 µC
Φ B = BA = Bπ r 2
The enclosed flux is
The particle moves according to
∑ F = ma:
qv B sin 90° =
r=
mv
qB
ΦB =
Then
(15 × 10
−6
mv 2
r
Bπ m 2 v 2
q2 B2
T ⋅ m 2 ) ( 30 × 10 −9 C) ( 0.6 T)
2
(a)
ΦB q2 B
v=
=
π m2
(b)
Energy for the particle-electric field system is conserved in the firing process:
Ui = K f :
q∆V =
π ( 2 × 10
−16
kg )
2
1
mv 2
2
−16
5
m v 2 ( 2 × 10 kg ) ( 2.54 × 10 m s )
∆V =
=
= 215 V
2q
2 ( 30 × 10 −9 C )
2
= 2.54 × 10 5 m s
Faraday’s Law
P31.55
(a)
ε = Bv = 0.360 V
(b)
FB = I B = 0.108 N
(c)
I=
ε=
R
219
0.900 A
Since the magnetic flux B ⋅ A is in effect decreasing, the
induced current flow through R is
from b to a. Point b is at higher potential.
FIG. P31.55
(d)
P31.56
No . Magnetic flux will increase through a loop to the
left of ab. Here counterclockwise current will flow to
produce upward magnetic field. The current in R is still from b to a.
ε = Bv at a distance r from wire
ε
⎛µ I ⎞
= ⎜ 0 ⎟ v
⎝ 2π r ⎠
I
FIG. P31.56
P31.57
ε = − d ( NBA) = −1⎛⎜⎝ dB ⎞⎟⎠ π a 2 = π a 2 K
dt
dt
(a)
Q = Cε = C π a 2 K
(b)
B into the paper is decreasing; therefore, current will attempt to counteract this. Positive
charge will go to upper plate .
(c)
*P31.58 (a)
(b)
The changing magnetic field through the enclosed area induces an electric field ,
surrounding the B -field, and this pushes on charges in the wire.
We would need to know whether the field is increasing or decreasing.
⎛
⎝
P =εI =ε2/R = ⎜N
⎛ dB 2 ⎞
R = ⎜N
πr ⎟
⎠
⎝ dt
P31.59
2
⎞
dB 2
π r cos 0° ⎟
⎠
dt
2
R
2
⎡220(0.020 T/s)π (0.12 m)2 ⎤⎦
P=⎣
= 248 mΩ
160 W
Higher resistance would reduce the power delivered.
The flux through the coil is Φ B = B ⋅ A = BA cos θ = BA cos ω t . The induced emf is
ε = − N d Φ B = − NBA d ( cos ω t ) = NBAω sin ω t
dt
(a)
(b)
(c)
dt
εmax = NBAω = 60.0 (1.00 T) (0.100 × 0.200 m 2 ) (30.0 rad s) =
dΦ B ε
dΦ B
ε
36.0 V
= , thus
= max =
= 0.600 V = 0.600
dt
N
dt
max
N
60.0
Wb s
At t = 0.050 0 s, ω t = 1.50 rad and
ε = ε max sin (1.50 rad ) = (36.0 V ) sin (1.50 rad ) =
(d)
36.0 V
35.9 V
The torque on the coil at any time is in magnitude
ε
⎛ε
τ = µ × B = NIA × B = ( NAB ) I sin ω t = ⎜ max ⎞⎟ ⎛ ⎞ sin ω t
⎝ ω ⎠ ⎝ R⎠
When
ε = εmax ,
sin ω t = 1.00 and τ =
2
εmax
=
ωR
(36.0 V)
= 4.32 N ⋅ m .
30
.
0
rad s)(10.0 Ω)
(
2
220
P31.60
Chapter 31
(a)
We use
ε = −N ∆Φ B , with
∆t
N = 1.
Taking a = 5.00 × 10 −3 m to be the radius of the washer, and h = 0.500 m,
⎛ µ0 I
µ I⎞
∆Φ B = B2 A − B1 A = A ( B2 − B1 ) = π a 2 ⎜
− 0
⎝ 2π ( h + a ) 2π a ⎟⎠
=
− µ0 ahI
a 2 µ0 I ⎛ 1
1
− ⎞=
2 ⎝ h + a a ⎠ 2(h + a)
2h
.
g
The time for the washer to drop a distance h (from rest) is: ∆t =
Therefore,
and
ε=(
ε=
µ 0 ahI
µ ahI
µ 0 aI
g
gh
= 0
=
2 ( h + a ) ∆t 2 ( h + a ) 2 h 2 ( h + a ) 2
4π × 10 −7 T ⋅ m A ) ( 5.00 × 10 −3 m ) (10.0 A )
2 ( 0.500 m + 0.005 00 m )
(9.80 m s ) ( 0.500 m )
2
2
= 97.4 nV
(b)
P31.61
Since the magnetic flux going through the washer (into the plane of the paper) is
decreasing in time, a current will form in the washer so as to oppose that decrease.
Therefore, the current will flow in a clockwise direction .
Find an expression for the flux through a rectangular area “swept
out” by the bar in time t. The magnetic field at a distance x from
wire is
B=
v
r
µ0 I
and Φ B = ∫ BdA. Therefore,
2π x
µ 0 I vt
2π
time t.
ΦB =
Then,
P31.62
I
ε
=
r +
∫
r
dx where vt is the distance the bar has moved in
x
FIG. P31.61
µ Iv ⎛ ⎞
dΦ B
= 0 ln ⎜1 + ⎟ .
⎝ r⎠
dt
2π
µ I
The magnetic field at a distance x from a long wire is B = 0 . We find an expression for the
2
πx
flux through the loop.
dΦ B =
µ0 I
(dx )
2π x
Therefore,
so
µ I
ΦB = 0
2π
ε = − dΦ
dt
B
r +w
∫
r
=
dx µ 0 I ⎛ w ⎞
ln ⎜1 + ⎟
=
2π ⎝ r ⎠
x
ε µ I v w
µ 0 I v w
and I = = 0
2π Rr ( r + w )
R
2π r (r + w)
vt
Faraday’s Law
P31.63
P31.64
(
Φ B = 6.00t 3 − 18.0t 2 T ⋅ m 2
and
ε = − d Φ B = −18.0t 2 + 36.0t
Maximum e occurs when
dε
= −36.0t + 36.0 = 0
dt
which gives
t = 1.00 s
Therefore, the maximum current (at t = 1.00 s ) is
I=
dt
For the suspended mass, M:
For the sliding bar, m:
∑ F = Mg − T = Ma.
∑ F = T − I B = ma,
where I =
ε = ( −18.0 + 36.0 ) V =
3.00 Ω
R
6.00 A
ε = Bv
R
R
a=
dv
Mg
B 2 2 v
=
−
dt m + M R ( M + m )
dv
∫0 (α − β v ) = ∫0 dt where
α=
Mg
B2 2
and β =
M +m
R( M + m)
Therefore, the velocity varies with time as
v=
Bv
= ( m + M ) a or
R
2 2
v
P31.65
)
We are given
Mg −
t
2 2
α
MgR
⎡1 − e− B t R( M + m ) ⎤
1 − e− β t ) =
(
2 2 ⎣
⎦
β
B
Suppose the field is vertically down. When an electron is moving away
from you the force on it is in the direction given by
qv × Bc as − ( away ) × down = −
= − left = right.
Therefore, the electrons circulate clockwise.
(a)
As the downward field increases, an emf is induced to produce some
current that in turn produces an upward field. This current is
directed
FIG. P31.65
counterclockwise, carried by negative electrons moving clockwise.
Therefore the original electron motion speeds up.
(b)
221
At the circumference, we have
∑F
c
= mac:
q v Bc sin 90° =
mv 2
r
m v = q rBc
The increasing magnetic field Bav in the area enclosed by the orbit produces a tangential
electric field according to
d ∫ E ⋅ d s = − dt B
av
⋅A
E (2π r ) = π r 2
dBav
dt
An electron feels a tangential force according to
Then
and
q
r dBav
dv
=m
2 dt
dt
E=
∑ F = ma :
t
t
r dBav
2 dt
qE=m
q
dv
dt
r
Bav = mv = q rBc
2
Bav = 2 Bc
222
Chapter 31
*P31.66 (a)
The induced emf is
ε = Bv
where B =
µ0 I
, v f = vi + gt = ( 9.80 m s 2 ) t , and
2π y
1
y f = yi − gt 2 = 0.800 m − ( 4.90 m s 2 ) t 2
2
4π × 10 −7
ε= (
T ⋅ m A ) ( 200 A )
2π ⎡⎣ 0.800 m − ( 4.90 m s 2 ) t 2 ⎤⎦
P31.67
( 0.300 m ) ( 9.80 m s2 ) t =
(b)
The emf is zero at t = 0.
(c)
The emf diverges to infinity at 0.404 s.
(d)
At t = 0.300 s,
1.18 × 10 −4 ) ( 0.300 )
ε= (
2
⎡⎣ 0.800 − 4.90 ( 0.300 ) ⎤⎦
(1.18 × 10 ) t
−4
⎡⎣ 0.800 − 4.90t 2 ⎤⎦
V
V = 98.3 µ V .
The magnetic field produced by the current in the straight wire is perpendicular to the plane of the coil at all points within the coil. The
µ I
magnitude of the field is B = 0 . Thus, the flux linkage is
2π r
h+ w
µ0 NIL dr µ0 NI max L ⎛ h + w ⎞
=
ln
sin (ω t + φ )
⎝ h ⎠
2π ∫h r
2π
Finally, the induced emf is
NΦB =
ε = − µ0 NI max Lω ln ⎛⎜⎝1 + w ⎞⎟⎠ cos (ωt + φ )
ε =−(
2π
4 π × 10
FIG. P31.67
h
−7
) (100)(50.0)(0.200 m ) (200π s ) ln ⎛⎜1 + 5.00 cm ⎞⎟ cos (ωt + φ )
−1
2π
⎝
5.00 cm ⎠
ε = − (87.1 mV) cos (200π t + φ )
The term sin (ωt + φ ) in the expression for the current in the straight wire does not change appreciably when ωt changes by 0.10 rad or less. Thus, the current does not change appreciably during
a time interval
∆t <
0.10
= 1.6 × 10 −4 s
( 200π s−1 )
We define a critical length c∆t = (3.00 × 108 m s) (1.6 × 10 −4 s) = 4.8 × 10 4 m equal to the distance
to which field changes could be propagated during an interval of 1.6 × 10 −4 s. This length is so
much larger than any dimension of the coil or its distance from the wire that, although we consider the straight wire to be infinitely long, we can also safely ignore the field propagation effects
in the vicinity of the coil. Moreover, the phase angle can be considered to be constant along the
wire in the vicinity of the coil.
If the angular frequency w were much larger, say, 200π × 10 5 s −1 , the corresponding critical
length would be only 48 cm. In this situation propagation effects would be important and the
above expression for ε would require modification. As a general rule we can consider field
ω
propagation effects for circuits of laboratory size to be negligible for frequencies, f =
, that
2π
are less than about 10 6 Hz.
Faraday’s Law
223
ANSWERS TO EVEN PROBLEMS
P31.2
0.800 mA
P31.4
(a) If the meter has a sufficiently short response time, it will register the current. The average
current may even run the meter offscale by a factor of 4.42, so you might wish to slow down the
motion of the coil. (b) Positive. The coil sees decreasing external magnetic flux toward you, so
it makes some flux of its own in this direction by carrying counterclockwise current, that enters
the red terminal of the ammeter.
P31.6
–10.2 mV
P31.8
(a) (m0IL/2p) ln(1 + w/h) (b) –4.80 mV; current is counterclockwise
P31.10
(a)
P31.12
(a) 5.33 V (b) A flat compact circular coil with 130 turns and radius 40.0 cm carries current
3.00 A counterclockwise. The current is smoothly reversed to become 3.00 A clockwise after
13.0 ms. At the center of this primary coil is a secondary coil in the same plane, with 20 turns and
radius 3.00 cm. Find the emf induced in the secondary.
P31.14
61.8 mV
P31.16
(a) See the solution. (b) If the wire is not centered, the coil will respond to stronger magnetic
fields on one side, but to correspondingly weaker fields on the opposite side. The emf induced
in the coil is proportional to the line integral of the magnetic field around the circular axis of the
toroid. Ampère’s law says that this line integral depends only on the amount of current the coil
encloses. It does not depend on the shape or location of the current within the coil, or on any
currents outside the coil.
P31.18
(0.422 V)cos w t
P31.20
(a) eastward
P31.22
1.00 m s
P31.24
(a) 500 mA
(b) 2.00 W
P31.26
(a) 233 Hz
(b) 1.98 mV
P31.28
(a) to the right
P31.30
Negative; see the solution.
P31.32
(a) 8.00 × 10 −21 N downward perpendicular to r1
P31.34
(a) ( 9.87 mV m ) cos (100 π t )
P31.36
(a) (19.6 V) sin ( 314 t )
P31.38
See the solution.
2
2
µ 0 nπ r22 ∆I
counterclockwise (b) µ0 nπ r2 ∆I
2 R ∆t
4 r1 R ∆t
(c) left
(b) 458 mV
(c) 2.00 W
(d) They are physically identical.
(b) out of the plane of the paper
(b) clockwise
(b) 19.6 V
(c) to the right
(b) 1.33 s
(d) into the paper
224
Chapter 31
(a) 8.00 Wb cos(377t) (b) 3.02 V sin(377t)
(e) 24.1 mN·m sin2(377t)
P31.42
(a)
P31.44
(a) counterclockwise (b) 100 mA (c) The speed of the bar increases until the backward magnetic force exerted on the current in the bar is equal to 0.6 N. The terminal speed is 281 m/s
(d) 169 W (e) The terminal speed becomes larger. The bar must move faster to generate a larger
emf to produce enough current in the larger resistance to feel the 0.6-N magnetic force.
(f) The power delivered to the circuit by the agent moving the bar, and then converted into
internal energy by the resistor, is described by P = F v = F 2 R / B 2 2 . Thus the power is directly
proportional to the resistance.
P31.46
(a) 43.8 A
P31.48
62.3 mA down through 6.00 Ω, 860 mA down through 5.00 Ω, 923 mA up through 3.00 Ω
P31.50
See the solution.
P31.52
(a) See the solution.
P31.54
(a) 254 km/s
P31.56
See the solution.
P31.58
(a) We would need to know whether the field is increasing or decreasing.
resistance would reduce the power delivered.
P31.60
(a) 97.4 nV
P31.62
µ 0 I v w
2 π Rr ( r + w )
P31.64
MgR ⎡
− B 2 2 t R( M + m )⎤
1− e
⎦
B 2 2 ⎣
P31.66
(a) 118t mV/(0.8 − 4.9t2) where t is in seconds.
diverges to infinity at 0.404. (d) 98.3 mV
2
π Bmax
R 4 bω 2
16 ρ
(b) 4 times larger
(c) 3.02 A sin(377t)
(d) 9.10 Wsin2 (377t)
P31.40
(c) 4 times larger
(d) 16 times larger
(b) 38.3 W
(b) 0.250 T
(b) 215 V
(b) 248 m Ω. Higher
(b) clockwise
(b) The emf is zero at t = 0.
(c) The emf
32
Inductance
CHAPTER OUTLINE
32.1
32.2
32.3
32.4
32.5
32.6
Self-Induction and Inductance
RL Circuits
Energy in a Magnetic Field
Mutual Inductance
Oscillations in an LC Circuit
The RLC Circuit
ANSWERS TO QUESTIONS
Q32.1
The coil has an inductance regardless of the nature of the
current in the circuit. Inductance depends only on the coil
geometry and its construction. Since the current is constant, the self-induced emf in the coil is zero, and the coil
does not affect the steady-state current. (We assume the
resistance of the coil is negligible.)
Q32.2
The inductance of a coil is determined by (a) the geometry
of the coil and (b) the “contents” of the coil. This is similar
to the parameters that determine the capacitance of a
capacitor and the resistance of a resistor. With an inductor,
the most important factor in the geometry is the number
of turns of wire, or turns per unit length. By the “contents”
we refer to the material in which the inductor establishes
a magnetic field, notably the magnetic properties of the
core around which the wire is wrapped.
*Q32.3 The emf across an inductor is zero whenever the current is constant, large or small. Answer (d).
*Q32.4 The fine wire has considerable resistance, so a few seconds is many time constants. The final
current is not affected by the inductance of the coil. Answer (c).
*Q32.5 The inductance of a solenoid is proportional to the number of turns squared, to the cross-sectional
area, and to the reciprocal of the length. Coil A has twice as many turns with the same length of
wire, so its circumference must be half as large as that of coil B. Its radius is half as large and its
area one quarter as large. For coil A the inductance will be different by the factor 22(1/4)(1/2) =
1/2. Answer (e).
Q32.6
When it is being opened. When the switch is initially standing open, there is no current in the
circuit. Just after the switch is then closed, the inductor tends to maintain the zero-current
condition, and there is very little chance of sparking. When the switch is standing closed, there
is current in the circuit. When the switch is then opened, the current rapidly decreases. The
induced emf is created in the inductor, and this emf tends to maintain the original current.
Sparking occurs as the current bridges the air gap between the contacts of the switch.
*Q32.7 Just before the switch is thrown, the voltage across the twelve-ohm resistor is very nearly 12 V.
Just after the switch is thrown, the current is nearly the same, maintained by the inductor. The
voltage across the 1 200-Ω resistor is then much more than 12 V. By Kirchhoff's loop rule,
the voltage across the coil is larger still: ∆VL > ∆V1 200 Ω > 12.0 V > ∆V12 Ω.
225
226
Chapter 32
*Q32.8 (i) (a) The bulb glows brightly right away, and then more and more faintly as the capacitor
charges up. (b) The bulb gradually gets brighter and brighter, changing rapidly at first and then
more and more slowly. (c) The bulb gradually gets brighter and brighter. (d) The bulb
glows brightly right away, and then more and more faintly as the inductor starts carrying more
and more current.
(ii) (a) The bulb goes out immediately. (b) The bulb glows for a moment as a spark jumps
across the switch. (c) The bulb stays lit for a while, gradually getting fainter and fainter.
(d) The bulb suddenly glows brightly. Then its brightness decreases to zero, changing rapidly at
first and then more and more slowly.
*Q32.9 The wire’s magnetic field goes in circles around it. We want this field to “shine” perpendicularly
through the area of the coil. Answer (c).
Q32.10 A physicist’s list of constituents of the universe in 1829 might include matter, light, heat, the stuff
of stars, charge, momentum, and several other entries. Our list today might include the quarks,
electrons, muons, tauons, and neutrinos of matter; gravitons of gravitational fields; photons of
electric and magnetic fields; W and Z particles; gluons; energy; momentum; angular momentum;
charge; baryon number; three different lepton numbers; upness; downness; strangeness; charm;
topness; and bottomness. Alternatively, the relativistic interconvertibility of mass and energy,
and of electric and magnetic fields, can be used to make the list look shorter. Some might think
of the conserved quantities energy, momentum, … bottomness as properties of matter, rather than
as things with their own existence. The idea of a field is not due to Henry, but rather to Faraday,
to whom Henry personally demonstrated self-induction. Still the thesis stated in the question has
an important germ of truth. Henry precipitated a basic change if he did not cause it. The biggest
difference between the two lists is that the 1829 list does not include fields and today’s list does.
*Q32.11 The energy stored in the magnetic field of an inductor is proportional to the square of the current.
1
Doubling I makes U = LI 2 get four times larger. Answer (a).
2
*Q32.12 Cutting the number of turns in half makes the inductance four times smaller. Doubling the
current would by itself make the stored energy four times larger, to just compensate. Answer (b).
Q32.13
The energy stored in a capacitor is proportional to the square of the electric field, and the energy stored
in an induction coil is proportional to the square of the magnetic field. The capacitor’s energy is
proportional to its capacitance, which depends on its geometry and the dielectric material inside. The
coil’s energy is proportional to its inductance, which depends on its geometry and the core material. On the other hand, we can think of Henry’s discovery of self-inductance as fundamentally new.
Before a certain school vacation at the Albany Academy about 1830, one could visualize the universe
as consisting of only one thing, matter. All the forms of energy then known (kinetic, gravitational,
elastic, internal, electrical) belonged to chunks of matter. But the energy that temporarily maintains a
current in a coil after the battery is removed is not energy that belongs to any bit of matter. This energy
is vastly larger than the kinetic energy of the drifting electrons in the wires. This energy belongs to
the magnetic field around the coil. Beginning in 1830, Nature has forced us to admit that the universe
consists of matter and also of fields, massless and invisible, known only by their effects.
Inductance
*Q32.14 (a)
The instant after the switch is closed, the situation is
as shown in the circuit diagram of Figure (a).
The requested quantities are:
I L = 0, I C =
ε0 , I
R
R
=
IL = 0
–
∆VL = e 0
IR = e 0 /R
ε0
Q=0
∆VC = 0
R
∆VR = e 0
+
∆VL = ε 0 , ∆VC = 0, ∆VR = ε 0
(b)
+
227
IC = e 0 /R
e 0–
Figure (a)
After the switch has been closed a long time, the
steady-state conditions shown in Figure (b) will
exist. The currents and voltages are:
– ∆VL = 0
IL = 0 +
IR = 0
I L = 0, I C = 0, I R = 0
Q = Ce 0
∆VL = 0, ∆VC = ε 0 , ∆VR = 0
∆VC = e 0
∆VR = 0
+
e 0–
Figure (b)
FIG. Q32.14
*Q32.15 (i) Answer (a). The mutual inductance of two loops in free space—that is, ignoring the use of
cores—is a maximum if the loops are coaxial. In this way, the maximum flux of the primary loop
will pass through the secondary loop, generating the largest possible emf given the changing
magnetic field due to the first.
(ii) Answer (c). The mutual inductance is a minimum if the magnetic field of the first coil lies in
the plane of the second coil, producing no flux through the area the second coil encloses.
Q32.16 When the capacitor is fully discharged, the current in the circuit is a maximum. The inductance of
the coil is making the current continue to flow. At this time the magnetic field of the coil contains
all the energy that was originally stored in the charged capacitor. The current has just finished
discharging the capacitor and is proceeding to charge it up again with the opposite polarity.
4L
4L
, then the oscillator is overdamped—it will not oscillate. If R <
, then the
C
C
oscillator is underdamped and can go through several cycles of oscillation before the radiated
signal falls below background noise.
Q32.17 If R >
Q32.18 An object cannot exert a net force on itself. An object cannot create momentum out of nothing.
A coil can induce an emf in itself. When it does so, the actual forces acting on charges in
different parts of the loop add as vectors to zero. The term electromotive force does not refer to
a force, but to a voltage.
SOLUTIONS TO PROBLEMS
Section 32.1
Self-Induction and Inductance
P32.1
ε = − L ∆I = ( −2.00 H ) ⎛⎜ 0 − 0.500 A ⎞⎟ ⎛⎝ 1 V ⋅ s ⎞⎠ =
P32.2
Treating the telephone cord as a solenoid, we have:
∆t
⎝ 0.010 0 s ⎠ 1 H ⋅ A
100 V
2
−7
−3
µ N 2 A ( 4π × 10 T ⋅ m A ) ( 70.0 ) π ( 6.50 × 10 m )
L= 0
=
= 1.36 µ H
0.600 m
2
228
*P32.3
Chapter 32
ε = − L dI = − L d ( I max sin ω t ) = − Lω I max cos ω t = − (10.0 × 10−3 ) (120π )(5.00) cos ω t
dt
dt
ε = − (6.00π ) cos (120π t ) = − (18.8 V) cos ( 377t )
P32.4
ε
From
∆I
= L ⎛ ⎞ , we have
⎝ ∆t ⎠
L=
NΦB
From L =
, we have
I
P32.5
P32.6
( ∆I
ΦB =
∆t )
=
24.0 × 10 −3 V
= 2.40 × 10 −3 H
10.0 A s
−3
LI ( 2.40 × 10 H ) ( 4.00 A )
= 19.2 µ T ⋅ m 2
=
500
N
2
−4
µ0 N 2 A µ0 ( 420 ) ( 3.00 × 10 )
=
= 4.16 × 10 −4 H
0.160
−6
ε = − L dI → dI = −ε = −175 × 10−4 V = −0.421 A s
dt
dt L 4.16 × 10 H
L=
ε
=L
dI
d
= ( 90.0 × 10 −3 ) (t 2 − 6t ) V
dt
dt
(a)
At t = 1.00 s,
ε=
360 mV
(b)
At t = 4.00 s,
ε=
180 mV
(c)
ε = (90.0 × 10−3 ) (2t − 6) = 0
t = 3.00 s
when
P32.7
ε
(a)
450 ⎞
B = µ0 nI = µ0 ⎛
( 0.040 0 A ) = 188 µT
⎝ 0.120 ⎠
(b)
Φ B = BA = 3.33 × 10 −8 T ⋅ m 2
(c)
L=
(d)
NΦB
= 0.375 mH
I
B and Φ B are proportional to current; L is independent of current.
µ N 2A
N Φ B NBA NA µ0 NI
=
≈
⋅
= 0
2π R
I
I
I 2π R
P32.8
L=
P32.9
ε = ε0 e− kt = − L dI
dt
FIG. P32.8
ε
dI = − 0 e− kt dt
L
If we require I → 0 as t → ∞, the solution is
Q = ∫ Idt =
ε
∫ kL e
∞
0
0
− kt
dt = −
ε0
2
k L
I=
ε0 e− kt = dq
kL
Q =
dt
ε0
k2L
Inductance
Section 32.2
P32.10
RL Circuits
Taking τ =
IR + L
L
, I = I i e−t τ :
R
dI
= 0 will be true if
dt
Because τ =
P32.11
(a)
⎛ 1⎞
dI
= I i e−t τ ⎜− ⎟
⎝ τ⎠
dt
1
I i R e− t τ + L ( I i e− t τ ) ⎛ − ⎞ = 0
⎝ τ⎠
L
, we have agreement with 0 = 0.
R
At time t,
I (t ) =
where
τ=
After a long time,
I max
ε (1 − e−t τ )
I (A)
Imax
R
1
L
= 0.200 s
R
0.5
ε ( 1 − e−∞) ε
=
=
R
t (s)
R
0
ε = ε (1 − e−t 0.200 s )
At I ( t ) = 0.500 I max
(0.500)
so
0.500 = 1 − e− t 0.200 s
Isolating the constants
on the right,
(b)
229
R
0
0.2
0.4
0.6
FIG. P32.11
R
ln ( e− t 0.200 s ) = ln ( 0.500 )
t
= −0.693
0.200 s
and solving for t,
−
or
t = 0.139 s
Similarly, to reach 90% of I max ,
0.900 = 1 − e− t τ
and
t = −τ ln (1 − 0.900 )
Thus,
t = − ( 0.200 s ) ln ( 0.100 ) = 0.461 s
*P32.12 The current increases from 0 to asymptotically
approach 500 mA. In case (a) the current jumps up
essentially instantaneously. In case (b) it increases
with a longer time constant, and in case (c) the increase
is still slower.
I
500 mA
(a)
(b)
(c)
0
t
0
FIG. P32.12
230
P32.13
Chapter 32
τ=
(b)
6.00 V ⎞
1 − e−0.250 2.00 ) = 0.176 A
I = I max (1 − e− t τ ) = ⎛
⎝ 4.00 Ω ⎠ (
(c)
I max =
I=
ε = 6.00 V =
R
4.00 Ω
0.800 = 1 − e
(d)
P32.14
L
= 2.00 × 10 −3 s = 2.00 ms
R
(a)
ε
− t 2.00 ms
1.50 A
FIG. P32.13
→ t = − ( 2.00 ms ) ln ( 0.200 ) = 3.22 ms
120
1− e ) =
(
(1 − e
R
9.00
−t τ
−1.80 7.00
) = 3.02 A
∆VR = IR = ( 3.02 ) ( 9.00 ) = 27.2 V
∆VL = ε − ∆VR = 120 − 27.2 = 92.8 V
P32.15
(a)
(b)
∆VR = IR = (8.00 Ω ) ( 2.00 A ) = 16.0 V
and
∆VL = ε − ∆VR = 36.0 V − 16.0 V = 20.0 V
Therefore,
∆VR 16.0 V
=
= 0.800
∆VL 20.0 V
∆VR = IR = ( 4.50 A ) (8.00 Ω ) = 36.0 V
FIG. P32.15
∆VL = ε − ∆VR = 0
P32.16
After a long time, 12.0 V = ( 0.200 A ) R. Thus, R = 60.0 Ω. Now, τ =
L
gives
R
L = τ R = ( 5.00 × 10 −4 s ) ( 60.0 V A ) = 30.0 mH
P32.17
τ=
P32.18
dI
1
= − I max ( e− t τ ) ⎛ − ⎞
⎝ τ⎠
dt
I = I max (1 − e−t τ ) :
L 15.0 H
=
= 0.500 s:
R 30.0 Ω
dI R
= I max e− t τ
dt L
and
I max =
ε
R
dI R
ε 100 V = 6.67 A s
= I max e0 = =
dt L
L 15.0 H
(a)
t = 0:
(b)
t = 1.50 s:
dI ε −t τ
−1.50 ( 0.500 )
= e = ( 6.67 A s) e
= ( 6.67 A s) e−3.00 = 0.332 A s
dt L
Name the currents as shown. By Kirchhoff’s laws:
I1 = I 2 + I 3
(1)
+10.0 V − 4.00 I1 − 4.00 I 2 = 0
(2)
+10.0 V − 4.00 I1 − 8.00 I 3 − (1.00 )
dI 3
=0
dt
(3)
FIG. P32.18
continued on next page
Inductance
From (1) and (2),
+10.0 − 4.00 I1 − 4.00 I1 + 4.00 I 3 = 0
and
I1 = 0.500 I 3 + 1.25 A
Then (3) becomes
10.0 V − 4.00 ( 0.500 I 3 + 1.25 A ) − 8.00 I 3 − (1.00 )
dI 3
=0
dt
(1.00 H ) ⎛
dI 3 ⎞
+ (10.0 Ω ) I 3 = 5.00 V
⎝ dt ⎠
We solve the differential equation using equations from the chapter text:
⎛ 5.00 V ⎞ ⎡
−(10.0 Ω)t 1.00 H ⎤
I 3 (t ) = ⎜
⎟ 1− e
⎦=
⎝ 10.0 Ω ⎠ ⎣
(0.500 A) ⎡⎣1 − e−10 t s ⎤⎦
I1 = 1.25 + 0.500 I 3 = 1.50 A − ( 0.250 A) e−10 t s
P32.19
P32.20
L
L
3.00 H
, we get R =
=
= 1.00 × 10 3 Ω = 1.00 kΩ .
R
C
3.00 × 10 −6 F
(a)
Using τ = RC =
(b)
τ = RC = (1.00 × 10 3 Ω ) ( 3.00 × 10 −6 F ) = 3.00 × 10 −3 s = 3.00 ms
For t ≤ 0, the current in the inductor is zero . At t = 0, it starts to
grow from zero toward 10.0 A with time constant
τ=
L (10.0 mH )
=
= 1.00 × 10 −4 s
R
(100 Ω )
For 0 ≤ t ≤ 200 µs, I = I max (1 − e−t τ ) =
(10.0 A) (1 − e−10 000 t s )
.
At t = 200 µs, I = (10.00 A ) (1 − e−2.00 ) = 8.65 A.
Thereafter, it decays exponentially as I = I i e
I = ( 8.65 A) e
P32.21
τ=
−10 000(t −200 µs) s
−t ′ τ
FIG. P32.20
, so for t ≥ 200 µs,
= ( 8.65 A) e−10 000 t s+2.00 = ( 8.65 e2.00 A) e−10 000 t s =
(63.9 A) e−10 000t s
L 0.140
=
= 28.6 ms
R 4.90
I max =
ε = 6.00 V = 1.22 A
(a)
I = I max (1 − e− t τ )
R
4.90 Ω
so
0.220 = 1.22 (1 − e− t τ )
t = −τ ln ( 0.820 ) = 5.66 ms
e−t τ = 0.820 :
(b)
I = I max (1 − e−10.0 0.028 6 ) = (1.22 A ) (1 − e−350 ) = 1.22 A
(c)
I = I max e− t τ
and
0.160 = 1.22e− t τ
so
t = −τ ln ( 0.131) = 58.1 ms
FIG. P32.21
231
232
P32.22
Chapter 32
(a)
(b)
dI
For a series connection, both inductors carry equal currents at every instant, so
is the
dt
same for both. The voltage across the pair is
Leq
dI
dI
dI
= L1 + L2
dt
dt
dt
so
Leq
dI
dI
dI
= L1 1 = L2 2 = ∆VL
dt
dt
dt
where
Thus,
(c)
Leq
∆VL ∆VL ∆VL
=
+
Leq
L1
L2
and
I = I1 + I 2 and
dI dI1 dI 2
=
+
dt dt
dt
1
1
1
= +
Leq L1 L2
dI
dI
dI
+ Req I = L1 + IR1 + L2 + IR2
dt
dt
dt
Now I and
dI
are separate quantities under our control, so functional equality requires
dt
both Leq = L1 + L2
(d)
Leq = L1 + L2
∆V = Leq
and Req = R1 + R2 .
dI
dI
dI
dI dI1 dI 2
=
+
.
+ Req I = L1 1 + R1 I1 = L2 2 + R2 I 2 where I = I1 + I 2 and
dt
dt
dt
dt dt
dt
We may choose to keep the currents constant in time. Then,
1
1
1
=
+
Req R1 R2
We may choose to make the current swing through 0. Then,
1
1
1
= +
Leq L1 L2
This equivalent coil with resistance will be equivalent to the pair of real inductorss for
all other currents as well.
Section 32.3
P32.23
(
)
2
(68.0 )2 ⎡⎣π 0.600 × 10 −2 ⎤⎦
N2A
= 8.21 µ H
L = µ0
= µ0
0.080 0
U=
P32.24
Energy in a Magnetic Field
(a)
(
)
1 2 1
2
LI = 8.21 × 10 −6 H ( 0.770 A ) = 2.44 µ J
2
2
The magnetic energy density is given by
B2
( 4.50 T )
=
= 8.06 × 10 6 J m 3
2 µ0 2 (1.26 × 10 −6 T ⋅ m A )
2
u=
(b)
The magnetic energy stored in the field equals u times the volume of the solenoid
(the volume in which B is non-zero).
2
U = uV = ( 8.06 × 10 6 J m 3 ) ⎡⎣( 0.260 m ) π ( 0.031 0 m ) ⎤⎦ = 6.32 kJ
P32.25
u = ∈0
E2
= 44.2 nJ m 3
2
u=
B2
= 995 µ J m 3
2 µ0
Inductance
∞
P32.26
∫e
−2 Rt L
dt = −
0
*P32.27 (a)
L
2R
∞
∫e
0
−2 Rt L
233
⎛ −2 Rdt ⎞
L −∞ 0
L
L
L −2 Rt L ∞
e −e )=
e
=−
⎜
⎟=−
(0 − 1) =
(
0
⎝ L ⎠
2R
2R
2R
2R
P = I∆V = 3A 22 V = 66.0 W
(b)
P = I∆VR = I2R = (3A)2 5 Ω = 45.0 W
(c)
When the current is 3.00 A, Kirchhoff’s loop rule reads
+22.0 V − ( 3.00 A)( 5.00 Ω) − ∆VL = 0.
Then
FIG. P32.27
∆VL = 7.00 V
The power being stored in the inductor is
I ∆VL = ( 3.00 A ) ( 7.00 V ) = 21.0 W
(d)
P32.28
At all instants after the connection is made, the battery power is equal to the sum of the power
delivered to the resistor and the power delivered to the magnetic field. Just after t = 0 the
resistor power is nearly zero, and the battery power is nearly all going into the magnetic field.
Long after the connection is made, the magnetic field is absorbing no more power and the
battery power is going into the resistor.
From the equation derived in the text,
I=
ε
(a)
I=
ε = 2.00 A
The maximum current, after a long time t, is
At that time, the inductor is fully energized and
(b)
Plost = I 2 R = ( 2.00 A )2 ( 5.00 Ω ) = 20.0 W
(c)
Pinductor = I ( ∆Vdrop ) = 0
(d)
U=
R
(1 − e
− Rt L
)
R
P = I ( ∆V ) = ( 2.00 A ) (10.0 V ) = 20.0 W .
LI 2 (10.0 H )( 2.00 A )
=
= 20.0 J
2
2
2
P32.29
The total magnetic energy is the volume integral of the energy density, u =
B2
.
2 µ0
2
⎛ B 2 ⎞ ⎛ R ⎞4
⎛R⎞
Because B changes with position, u is not constant. For B = B0 ⎜ ⎟ , u = ⎜ 0 ⎟ ⎜ ⎟ .
⎝r⎠
⎝ 2µ 0 ⎠ ⎝ r ⎠
Next, we set up an expression for the magnetic energy in a spherical shell of radius r and
thickness dr. Such a shell has a volume 4 π r 2 dr, so the energy stored in it is
⎛ 2π B02 R 4 ⎞ dr
dU = u ( 4π r 2 dr ) = ⎜
⎝ µ0 ⎟⎠ r 2
We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the
sphere. This gives
2π ( 5.00 × 10 −5 T ) ( 6.00 × 10 6 m )
2π B02 R3
U=
=
= 2.70 × 1018 J
µ0
(1.26 × 10 −6 T ⋅ m A )
2
3
234
Chapter 32
Section 32.4
P32.30
Mutual Inductance
I1 (t ) = I max e−α t sin ω t with I max = 5.00 A, α = 0.025 0 s⫺1 , and ω = 377 rad s
dI1
= I max e−α t ( −α sin ω t + ω cos ω t )
dt
dI1
= ( 5.00 A s ) e− 0.020 0 ⎡⎣ − ( 0.025 0 ) sin ( 0.800 ( 377 ))
dt
At t = 0.800 s,
+ 377 cos ( 0.800 ( 377 )) ⎤⎦
Thus,
P32.31
ε 2 = − M dI1 :
dt
ε2 = − M dI1 = − (1.00 × 10−4 H ) (1.00 × 10 4 A s) cos (1 000t )
dt
(ε2 )max =
P32.32
dI1
= 1.85 × 10 3 A s
dt
−ε 2
+3.20 V
M=
=
= 1.73 mH
dI1 dt 1.85 × 10 3 A s
1.00 V
Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at
µ I
distance x from the wire is B = 0 , and this field passes perpendicularly through the plane
2π x
of the loop. The flux through the loop is
70 mm
1.7
µ I
dx µ0 I ⎛ 1.70 ⎞
ln
Φ B = ∫ B ⋅ dA = ∫ BdA = ∫ B ( dx ) = 0
=
⎝ 0.400 ⎠
2π 0.400∫ mm x
2π
The mutual inductance between the wire and the loop is then
M=
1( 4 π × 10 −7 T ⋅ m A) ( 2.70 × 10 −3 m )
N 2 Φ12 N 2 µ 0 I ⎛ 1.70 ⎞ N 2 µ 0 ln ⎜
=
⎟=
(1.45) =
(1.45)
⎝ 0.400 ⎠
2π I
2π
I1
2π
M = 7.81 × 10 −10 H = 781 pH
P32.33
P32.34
(
)
(a)
M=
−6
N B Φ BA 700 90.0 × 10
=
= 18.0 mH
IA
3.50
(b)
LA =
−6
Φ A 400 300 × 10
=
= 34.3 mH
3.50
IA
(c)
ε B = − M dI A = − (18.0 mH )(0.500 A/s) =
(
dt
)
−9.00 mV
(a)
Solenoid 1 creates nearly uniform field everywhere inside it, given by m0N1I/
The flux though one turn of solenoid 2 is m0p R22N1I/
The emf induced in solenoid 2 is –(m0p R22N1N2/)(dI/dt)
The mutual inductance is m0p R22N1N2/
(b)
Solenoid 2 creates nearly uniform field everywhere inside it, given by m0N2I/
and nearly zero field outside.
The flux though one turn of solenoid 1 is m0 p R22N2 I/
The emf induced in solenoid 1 is –(m0 p R22N1N2/)(dI/dt)
The mutual inductance is m0 p R22N1N2/
(c)
The mutual inductances are the same. This is one example of von Neumann’s rule,
mentioned in the next problem.
Inductance
P32.35
The large coil produces this field at the center of the small coil:
N1µ 0 I1 R12
3 2 . The field is
2 ( x 2 + R12 )
normal to the area of the small coil and nearly uniform over this area, so it produces flux
Φ12 =
N1µ 0 I1 R12
π R22 through the face area of the small coil. When current I1 varies,
2 (x + R )
this is the emf induced in the small coil:
2
ε2 = − N2
P32.36
2 32
1
N1 N 2πµ0 R12 R22 dI1
N1 N 2πµ0 R12 R22
dI1
d N1 µ0 R12π R22
M
I
=
−
=
−
so
M
=
32
32
32 1
dt
dt
dt 2 ( x 2 + R12 )
2 ( x 2 + R12 )
2 ( x 2 + R12 )
With I = I1 + I 2 , the voltage across the pair is:
∆V = − L1
dI1
dI
dI
dI
dI
− M 2 = − L2 2 − M 1 = − Leq
dt
dt
dt
dt
dt
−
So,
dI 2 M ( ∆V ) M 2 dI 2
+
+
= ∆V
dt
L1
L1 dt
( − L1 L2 + M 2 ) dIdt2 = ∆V ( L1 − M )
− L2
and
−
leads to
(− L L
Adding [1] to [2],
(− L L
So,
Leq = −
1
1
2
2
+ M2)
dI1
= ∆V ( L2 − M )
dt
+ M2)
dI
= ∆V ( L1 + L2 − 2 M )
dt
∆V
L1 L2 − M 2
=
dI dt
L1 + L2 − 2 M
Oscillations in an LC Circuit
1
1
2
At different times, (UC )max = (UL )max so ⎡⎢ C ( ∆V ) ⎤⎥ = ⎛ LI 2 ⎞
⎠ max
⎣2
⎦ max ⎝ 2
I max =
P32.38
FIG. P32.36
dI 2 ∆V M dI1
=
+
dt
L2 L2 dt
By substitution,
Section 32.5
P32.37
dI1 ∆V M dI 2
=
+
dt
L1 L1 dt
C
1.00 × 10 −6 F
( ∆V )max =
( 40.0 V ) = 0.400 A
L
10.0 × 10 −3 H
⎡ 1 C ( ∆V )2 ⎤ =
⎢⎣ 2
⎥⎦ max
20.0 × 10 −3 H
L
⎛ 1 LI 2 ⎞ so ∆V
=
I
=
(
)
(0.100 A) = 20.0 V
C max
max
⎝2
⎠ max
0.500 × 10 −6 F
C
235
236
P32.39
Chapter 32
When the switch has been closed for a long time, battery, resistor,
and coil carry constant current I max =
ε . When the switch is opened,
R
current in battery and resistor drops to zero, but the coil carries this
same current for a moment as oscillations begin in the LC loop.
We interpret the problem to mean that the voltage amplitude of these
1
1 2
2
oscillations is ∆V, in C ( ∆V ) = LI max
.
2
2
Then, L =
P32.40
C ( ∆V )
2
2
I max
1
C ( ∆V ) R 2
2
=
(0.500 × 10
F ) (150 V) ( 250 Ω)
−6
2
(50.0 V)
1
f=
(b)
Q = Qmax cos ω t = (180 µC) cos ( 847 × 0.001 00 ) = 119 µC
(c)
I=
(0.082 0 H ) (17.0 × 10−6
2π
F)
2
2
(a)
2 π LC
=
ε2
=
FIG. P32.39
= 0.281 H .
= 135 Hz
dQ
= −ω Qmax sin ω t = − ( 847 )(180 ) sin ( 0.847 ) = −1114 mA
dt
P32.41
This radio is a radiotelephone on a ship, according to frequency assignments made by
international treaties, laws, and decisions of the National Telecommunications and Information
Administration.
1
The resonance frequency is f0 =
2 π LC
1
1
C=
=
= 608 pF
Thus,
2
6
( 2π f0 ) L ⎡⎣ 2π (6.30 × 10 Hz )⎤⎦ 2 (1.05 × 10 −6 H )
P32.42
(a)
f=
(b)
Q = Cε = (1.00 × 10 −6 F ) (12.0 V) = 12.0 µC
(c)
1
1 2
Cε 2 = LI max
2
2
1
2 π LC
I max = ε
P32.43
=
(0.100 H ) (1.00 × 10−6
2π
F)
= 503 Hz
FIG. P32.42
C
1.00 × 10 −6 F
= 12 V
= 37.9 mA
L
0.100 H
(d)
At all times
ω=
1
=
LC
1
1
2
U = Cε 2 = (1.00 × 10 −6 F ) (12.0 V ) = 72.0 µ J
2
2
1
( 3.30 H ) (840 × 10−12 F )
Q = Qmax cos ω t , I =
(a)
1
(
= 1.899 × 10 4 rad s
dQ
= −ω Qmax sin ω t
dt
⎡⎣105 × 10 −6 ⎤⎦ cos ⎡⎣(1.899 × 10 4 rad s ) ( 2.00 × 10 −3 s ) ⎤⎦
Q2
UC =
=
2C
2 (840 × 10 −12 )
continued on next page
)
2
= 6.03 J
Inductance
(b)
UL =
UL =
(c)
237
Q 2 sin 2 (ω t )
1 2 1
2
LI = Lω 2Qmax
sin 2 (ω t ) = max
2
2
2C
(105 × 10
−6
C ) sin 2 ⎡⎣(1.899 × 10 4 rad s ) ( 2.00 × 10 −3 s ) ⎤⎦
= 0.529 J
2 (840 × 10 −12 F )
2
U total = U C + U L = 6.56 J
Section 32.6 The RLC Circuit
2
P32.44
(a)
ωd =
1 ⎛ R⎞
−
=
LC ⎝ 2 L ⎠
Therefore,
P32.45
fd =
2
⎛
⎞
7.60
−⎜
= 1.58 × 10 4 rad s
−3
−6
−3 ⎟
( 2.20 × 10 ) (1.80 × 10 ) ⎝ 2 ( 2.20 × 10 ) ⎠
1
ωd
= 2.51 kHz
2π
(b)
Rc =
4L
= 69.9 Ω
C
(a)
ω0 =
1
1
= 4.47 krad s
=
LC
( 0.500 ) ( 0.100 × 10 −6 )
(b)
ωd =
1 ⎛ R⎞
−
= 4.36 krad s
LC ⎝ 2 L ⎠
(c)
∆ω
= 2.53% lower
ω0
2
P32.46
Choose to call positive current clockwise in Figure 32.15. It drains charge from the capacitor
dQ
according to I = −
. A clockwise trip around the circuit then gives
dt
Q
dI
+ − IR − L = 0
C
dt
+
P32.47
Q dQ
d dQ
+
R+ L
= 0, identical with Equation 32.28.
C dt
dt dt
(a)
Q = Qmax e− Rt 2 L cos ω d t
so
I max ∝ e− Rt 2 L
0.500 = e−Rt 2 L
and
Rt
= − ln ( 0.500 )
2L
t =−
(b)
⎛ 2L ⎞
2L
ln ( 0.500 ) = 0.693⎜ ⎟
⎝ R⎠
R
2
U 0 ∝ Qmax
and U = 0.500U 0
t =−
so
⎛ 2L ⎞
2L
ln ( 0.707 ) = 0.347 ⎜ ⎟
⎝ R⎠
R
Q = 0.500Qmax = 0.707Qmax
(half as long)
238
Chapter 32
Additional Problems
P32.48
(a)
Let Q represent the magnitude of the opposite charges on the plates of a parallel plate
capacitor, the two plates having area A and separation d. The negative plate creates electric
Q
Q2
field E =
toward itself. It exerts on the positive plate force F =
toward the
2 ∈0 A
2 ∈0 A
Q
. The energy density is
negative plate. The total field between the plates is
∈0 A
1
1
Q2
Q2
u E = ∈0 E 2 = ∈0 2 2 =
. Modeling this as a negative or inward pressure,
2
2
∈0 A
2 ∈0 A 2
we have for the force on one plate F = PA =
(b)
Q2
, in agreement with our first analysis.
2 ∈0 A 2
The lower of the two current sheets shown creates
µ J
above it magnetic field B = 0 s −kˆ . Let and w
2
represent the length and width of each sheet. The
upper sheet carries current J s w and feels force
( )
Js
Js
y
x
µ J
µ wJ s2 ˆ
F = I × B = J s w 0 s ˆi × −kˆ = 0
j.
2
2
( )
z
FIG. P32.48(b)
µ J2
F
= 0 s .
The force per area is P =
2
w
(c)
µ0 J s ˆ µ0 J s ˆ
−k +
−k = µ 0 J s kˆ , with
2
2
magnitude B = µ 0 J s . Outside the space they enclose, the fields of the separate sheets are
Between the two sheets the total magnetic field is
( )
( )
in opposite directions and add to zero .
P32.49
µ J2
1 2 µ 02 J s2
B =
= 0 s
2µ 0
2µ 0
2
(d)
uB =
(e)
This energy density agrees with the magnetic pressure found in part (b).
(a)
ε L = − L dI = − (1.00 mH ) d (20.0t ) =
(b)
Q=
dt
dt
t
t
0
0
∫ Idt = ∫ (20.0t ) dt = 10.0t
∆VC =
−20.0 mV
2
−Q
−10.0t 2
=
= − (10.0 MV s 2 ) t 2
C 1.00 × 10 −6 F
(−10.0t 2 ) ≥ 1 1.00 × 10−3 (20.0t )2, then
Q2 1 2
≥ LI , or
When
(
)
2C 2
2 (1.00 × 10 −6 ) 2
2
(c)
100t 4 ≥ ( 400 × 10 −9 ) t 2 . The earliest time this is true is at t = 4.00 × 10 −9 s = 63.2 µs .
Inductance
P32.50
(a)
ε L = − L dI = − L d ( Kt ) =
(b)
I=
dt
dQ
,
dt
dt
− LK
Q=
so
When
t
0
0
1
2
−Q
Kt 2
= −
2C
C
1 ⎛ K 2t 4 ⎞ 1
2 2
C⎜
⎟ = L (K t )
2 ⎝ 4C 2 ⎠ 2
1
1
2
C ( ∆VC ) = LI 2 ,
2
2
t = 2 LC
Thus
1 Q2
1 ⎛Q ⎞ 1 2
=
⎜ ⎟ + LI
2 C 2C ⎝ 2 ⎠ 2
2
P32.51
t
∫ Idt = ∫ Ktdt = 2 Kt
∆VC =
and
(c)
239
so
The flux through each turn of the coil is
I=
3Q 2
4CL
ΦB =
LI
Q
=
N
2N
3L
C
where N is the number of turns.
*P32.52 (a)
(b)
(c)
(d)
The inductor has no voltage across it. It behaves as a short circuit . The battery sees
equivalent resistance 4 Ω + (1/4 Ω + 1/8 Ω)–1 = 6.67 Ω. The battery current is
10 V/6.67 Ω = 1.50 A. The voltage across the parallel combination of resistors is
10 V – 1.50 A 4 Ω = 4 V. The current in the 8-Ω resistor and the inductor is
4V/8Ω = 500 mA .
U = (1/2) LI 2 = (1/2) 1 H(0.5 A)2 = 125 mJ
The energy becomes 125 mJ of additional internal energy in the 8-Ω resistor and the 4-Ω
resistor in the middle branch.
The current decreases from 500 mA toward zero,
I
showing exponential decay with a time constant of
500 mA
L/R = 1 H/12 Ω = 83.3 ms.
0
t
0
83.3 ms
FIG. P32.52(d)
*P32.53 (a)
(b)
(c)
(d)
Just after the circuit is connected, the potential difference across the resistor is 0 and the
emf across the coil is 24.0 V.
After several seconds, the potential difference across the resistor is 24.0 V and that across
the coil is 0.
The resistor voltage and inductor voltage always add to 24 V. The resistor voltage increases
monotonically, so the two voltages are equal to each other, both being 12.0 V, just once.
The time is given by 12 V = IR = Rε /R(1 – e–Rt/L) = 24 V(1 – e–6Ωt/0.005 H).
This is 0.5 = e–1 200 t or 1 200 t = ln 2 giving t = 0.578 ms after the circuit is connected .
As the current decays the potential difference across the resistor is always equal to the emf
across the coil. It decreases from 24.0 V to zero.
240
Chapter 32
*P32.54 We have 9 V = 2 A R + L (0.5 A/s)
and
5 V = 2 A R + L (–0.5 A/s)
Solving simultaneously, 9 V – 5 V = L(1 A/s) so L = 4.00 H
*P32.55 Between t = 0 and t = 1 ms, the rate of change of current
is 2 A/s, so the induced voltage
∆Vab = –L dI/dt is –100 mV. Between t = 1 ms and
t = 2 ms, the induced voltage is zero. Between t = 2 ms
and t = 3 ms the induced voltage is –50 mV. Between
t = 3 ms and t = 5 ms, the rate of change of current is
(–3/2) A/s, and the induced voltage is +75 mV.
and 7 V = 2 A R so R = 3.50 Ω
∆Vab (mV)
100
0 0
2
4
6
t (ms)
−100
FIG. P32.55
*P32.56 (a)
ω = (LC)–1/2 = (0.032 V ⋅ s/A 0.000 5 C/V)–1/2 = 250 rad/s
(b)
2
2
⎛ 1
⎛
⎤ ⎞
⎡R⎤ ⎞
⎡
1
4Ω
−
ω = ⎜⎜
− ⎢ ⎥ ⎟⎟ = ⎜⎜
⎥ ⎟ = 242 rad/s
⎢⎣
−5 2
2 ⋅ 0.032 V ⋅ s/A ⎦ ⎟⎠
⎝ LC ⎣ 2 L ⎦ ⎠
⎝ 1.6 × 10 s
(c)
2
⎛ 1
⎡R⎤ ⎞
ω = ⎜⎜
− ⎢ ⎥ ⎟⎟
⎝ LC ⎣ 2 L ⎦ ⎠
1/ 2
1/ 2
1/ 2
2
⎛
⎤ ⎞
⎡
1
15 Ω
⎟
= ⎜⎜
−
⎥
⎢
−5 2
⎣ 2 ⋅ 0.032 V ⋅ s/A ⎦ ⎟⎠
⎝ 1.6 × 10 s
= 87.0 rad/s
2
2
⎛ 1
⎛
⎤ ⎞
⎡R⎤ ⎞
⎡
1
17 Ω
⎟ gives an imaginary answer. In
−
ω = ⎜⎜
− ⎢ ⎥ ⎟⎟ = ⎜⎜
⎥
⎢
−5 2
⎣ 2 ⋅ 0.032 V ⋅ s/A ⎦ ⎟⎠
⎝ LC ⎣ 2 L ⎦ ⎠
⎝ 1.6 × 10 s
parts (a), (b), and (c) the calculated angular frequency is experimentally verifiable.
1/ 2
(d)
1/ 2
1/ 2
Experimentally, in part (d) no oscillations occur . The circuit is overdamped.
*P32.57 B =
µ 0 NI
2π r
b
(a)
µ 0 NI
µ NIh
hdr = 0
2π
a 2π r
µ NIh ⎛ b ⎞
ln ⎜ ⎟
= 0
⎝a⎠
2π
ΦB =
∫ BdA = ∫
b
∫
a
dr
r
L=
µ N 2h ⎛ b ⎞
N ΦB
ln ⎜ ⎟
= 0
⎝a⎠
2π
I
L=
µ 0 ( 500 ) ( 0.010 0 ) ⎛ 12.0 ⎞
ln ⎜
⎟ = 91.2 µ H
⎝ 10.0 ⎠
2π
FIG. P32.57
2
(b)
(c)
2
µ 0 N 2 ⎛ A ⎞ µ 0 ( 500 ) ⎛ 2.00 × 10 −4 m 2 ⎞
⎜
⎟ = 90.9 µ H , This approximate result is
⎜ ⎟=
2π ⎝ R ⎠
2π
0.110
⎝
⎠
only 0.3% different from the precise result.
Lappx =
Inductance
P32.58
(a)
B=
At the center,
N µ0 IR 2
2( R + 0
2
)
2 32
=
241
N µ0 I
2R
N µ0 I
π
π R 2 cos 0° = N µ0 IR
2R
2
⎛π ⎞
dΦ B
dI
dI
= −N
≈ − N ⎜ ⎟ N µ0 R = − L
⎝2⎠
dt
dt
dt
So the coil creates flux through itself Φ B = BA cos θ =
When the current it carries changes,
εL
L≈
so
(b)
2 π r = 3 ( 0.3 m )
π 2
N µ0 R
2
r ≈ 0.14 m
so
π 2
(1 ) ( 4 π × 10−7 T ⋅ m A ) (0.14 m ) = 2.8 × 10−7 H
2
L ~ 100 nH
L≈
(c)
P32.59
L 2.8 × 10 −7 V ⋅ s A
=
= 1.0 × 10 −9 s
270 V A
R
L
~ 1 ns
R
Left-hand loop:
ε − ( I + I 2 ) R1 − I 2 R2 = 0
Outside loop:
ε − ( I + I 2 ) R1 − L dI = 0
dt
dI
ε ′ − IR ′ − L = 0
Eliminating I 2 gives
dt
This is of the same form as the differential equation 32.6 in
the chapter text for a simple RL circuit, so its solution is of the
same form as the equation 32.7 for the current in the circuit:
I (t ) =
But R′ =
R1 R2
and
R1 + R2
ε′ =
R′
(1 − e
R2ε
, so
R1 + R2
Thus
P32.60
ε′
− R ′t L
FIG. P32.59
)
ε ′ = ε R2 ( R1 + R2 ) = ε
R ′ R1 R2 ( R1 + R2 ) R1
ε 1 − e − R ′t L
I (t ) =
R1
(
)
From Ampère’s law, the magnetic field at distance r ≤ R is found as:
⎛ I ⎞
µ Ir
B (2π r ) = µ0 J (π r 2 ) = µ0 ⎜
π r 2 ) , or B = 0 2
2 ⎟(
2π R
⎝π R ⎠
The magnetic energy per unit length within the wire is then
R
R
µ I2
µ I2
µ I 2 ⎛ R4 ⎞
U
B2
=∫
( 2π rdr ) = 0 4 ∫ r 3 dr = 0 4 ⎜ ⎟ = 0
0 2 µ0
4π R 0
4π R ⎝ 4 ⎠
16π
This is independent of the radius of the wire.
242
P32.61
Chapter 32
(a)
While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of
the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around
the outer loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:
+ε 0 − ⎡⎣( 2.00 + 6.00 ) × 10 3 Ω ⎤⎦ ( 9.00 × 10 −3 A ) = 0
+ε 0 = 72.0 V with end b at the higher potential
(b)
FIG. P32.61(b)
(c)
After the switch is opened, the current around the outer loop decays as
I = I i e− Rt L with I max = 9.00 mA, R = 8.00 kΩ, and L = 0.400 H.
Thus, when the current has reached a value I = 2.00 mA, the elapsed time is:
9.00 ⎞
L
I
0.400 H ⎞
t = ⎛ ⎞ ln ⎛ i ⎞ = ⎛
ln ⎛
= 7.52 × 10 −5 s = 75.2 µs
3
⎝ R ⎠ ⎝ I ⎠ ⎝ 8.00 × 10 Ω ⎠
⎝ 2.00 ⎠
P32.62
(a)
(b)
2U
is non-zero.
I2
Every field line goes through the rectangle between the conductors.
It has a magnetic field, and it stores energy, so L =
w −a
(c)
Φ = LI so
Φ 1
L = = ∫ BdA
I I y=a
L=
L=
Thus
P32.63
1
I
w −a
∫
a
w− a
⎛µ I
µ0 I ⎞ 2 µ0 Ix
2 µ0 x
=
xdy ⎜ 0 +
=
ln
dy
y
2π
⎝ 2π y 2π ( w − y) ⎟⎠ I ∫ 2π y
a
µ0 x ⎛ w − a ⎞
ln
⎝ a ⎠
π
When the switch is closed, as shown
in Figure (a), the current in the
inductor is I:
12.0 − 7.50 I − 10.0 = 0 → I = 0.267 A
When the switch is opened, the initial
current in the inductor remains at
0.267 A.
IR = ∆V :
(0.267 A) R ≤ 80.0 V
R ≤ 300 Ω
P32.64
For an RL circuit,
I (t ) = I i e
−( R L )t
R
t = 10 −9
L
I (t )
:
Ii
= 1 − 10 −9 = e
FIG. P32.63
−( R L )t
Rmax =
R
t
L
(3.14 × 10 ) (10 ) =
( 2.50 yr ) ( 3.16 × 10 s yr )
−8
so
≅ 1−
−9
7
3.97 × 10 −25 Ω
(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm 2 , its
resistance would be at least 10 −6 Ω.)
Inductance
P32.65
243
2
1 2 1
LI = ( 50.0 H ) ( 50.0 × 10 3 A) = 6.25 × 1010 J
2
2
(a)
UB =
(b)
Two adjacent turns are parallel wires carrying current in the same direction. Since the loops
have such large radius, a one-meter section can be regarded as straight.
µ I
B= 0
Then one wire creates a field of
2π r
This causes a force on the next wire of F = I B sin θ
F = I
giving
µ0 I
µ I 2
sin 90 º = 0
2π r
2π r
(1.00 m ) ( 50.0 × 10 3 A )
F = ( 4π × 10 N A )
= 2 000 N
2π ( 0.250 m )
2
−7
Evaluating the force,
P32.66
2
1.00 × 10 9 W
= 5.00 × 10 3 A
∆V 200 × 10 3 V
µ I
From Ampère’s law, B ( 2 π r ) = µ 0 I enclosed or B = 0 enclosed
2π r
P = I ∆V
(a)
I=
P
=
At r = a = 0.020 0 m, I enclosed = 5.00 × 10 3 A
and B =
( 4π × 10
−7
T ⋅ m A ) ( 5.00 × 10 3 A )
2π ( 0.020 0 m )
FIG. P32.66
= 0.050 0 T = 50.0 mT
(b)
At r = b = 0.050 0 m , I enclosed = I = 5.00 × 10 3 A
B=
and
b
(c)
U = ∫ udV = ∫
−7
T ⋅ m A ) ( 5.00 × 10 3 A )
2π ( 0.050 0 m )
2
[ B ( r)] ( 2π rdr ) = µ0 I 2 −7
= 0.0020 0 T = 20.0 mT
b
2
dr µ0 I ⎛ b ⎞
ln
=
⎝ a⎠
4π ∫a r
4π
2 µ0
a
( 4π × 10
U=
( 4π × 10
T ⋅ m A ) ( 5.00 × 10 3 A ) (1 000 × 10 3 m )
2
4π
5.00 cm ⎞
ln ⎛
⎝ 2.00 cm ⎠
= 2.29 × 10 J = 2.29 MJ
6
(d)
The magnetic field created by the inner conductor exerts a force of repulsion on the current
in the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small
rectangular section of the outer cylinder of length and width w.
⎛
⎞
w
⎟
It carries a current of
(5.00 × 10 3 A) ⎜⎜ 2π (0.050
0 m ) ⎟⎠
⎝
and experiences an outward force
(5.00 × 10 A ) w 3
F = I B sin θ =
The pressure on it is
2π ( 0.050 0 m )
P=
( 20.0 × 10
−3
T ) sin 90.0°
3
−3
F F ( 5.00 × 10 A ) ( 20.0 × 10 T )
=
=
= 318 Pa
2π ( 0.050 0 m )
A w
244
P32.67
Chapter 32
−7
µ 0 NI ( 4 π × 10 T ⋅ m A) (1 400 )( 2.00 A)
=
= 2.93 × 10 −3 T ( upward )
1.20 m
(a)
B=
(b)
(2.93 × 10−3 T) = 3.42 J m 3 ⎛⎜1 N ⋅ m ⎞⎟
B2
u=
=
(
)⎝ 1 J ⎠
2µ 0 2 ( 4 π × 10 −7 T ⋅ m A)
2
= 3.42 N m 2 = 3.42 Pa
(c)
To produce a downward magnetic field, the surface of the superconductor must carry a
clockwise current.
(d)
(e)
The vertical component of the field of the solenoid exerts an inward force on the superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field diverges
and has a radially outward horizontal component. This component exerts upward force on
the clockwise superconductor current. The total force on the core is upward . You can
think of it as a force of repulsion between the solenoid with its north end pointing up, and
the core, with its north end pointing down.
2
F = PA = ( 3.42 Pa ) ⎡⎢π (1.10 × 10 −2 m ) ⎤⎥ = 1.30 × 10 −3 N
⎣
⎦
Note that we have not proved that energy density is pressure. In fact, it is not in some cases.
Chapter 21 proved that the pressure is two-thirds of the translational energy density in an
ideal gas.
ANSWERS TO EVEN PROBLEMS
P32.2
1.36 µ H
P32.4
19.2 µ Wb
P32.6
(a) 360 mV
P32.8
See the solution.
P32.10
See the solution.
P32.12
See the solution.
P32.14
92.8 V
P32.16
30.0 mH
P32.18
(500 mA)(1 – e−10t/s), 1.50 A – (0.250 A) e−10t/s
P32.20
0 for t < 0; (10 A)(1 – e−10 000t) for 0 < t < 200 µs; (63.9 A) e−10 000t for t > 200 µs
P32.22
(a), (b), and (c) See the solution.
P32.24
(a) 8.06 MJ m 3
P32.26
See the solution.
(b) 180 mV
(c) t = 3.00 s
(b) 6.32 kJ
(d) Yes; see the solution.
Inductance
(b) 20.0 W
P32.28
(a) 20.0 W
P32.30
1.73 mH
P32.32
781 pH
P32.34
(a) and (b) m0p R22N1N2/
P32.36
(L1L2 − M 2)/(L1 + L2 − 2M)
P32.38
20.0 V
P32.40
(a) 135 Hz
P32.42
(a) 503 Hz
P32.44
(a) 2.51 kHz
P32.46
See the solution.
P32.48
(b) m 0Js2/2 away from the other sheet
P32.50
(a)
P32.52
(a) a short circuit; 500 mA (b) 125 mJ (c) The energy becomes 125 mJ of additional internal
energy in the 8-Ω resistor and the 4-Ω resistor in the middle branch. (d) See the solution. The
current decreases from 500 mA toward zero, showing exponential decay with a time constant of
83.3 ms.
P32.54
L = 4.00 H and R = 3.50 Ω
P32.56
(a) 250 rad/s (b) 242 rad/s (c) 87.0 rad/s (d) In parts (a), (b), and (c) the calculated angular
frequency is experimentally verifiable. In part (d) the equation for w gives an imaginary answer.
Experimentally, no oscillations occur when the circuit is overdamped.
P32.58
(a) L ≈ (p /2)N 2m 0R
P32.60
See the solution.
P32.62
(a) It creates a magnetic field. (b) The long narrow rectangular area between the conductors
encloses all of the magnetic flux.
P32.64
3.97 × 10 −25 Ω
P32.66
(a) 50.0 mT (b) 20.0 mT (c) 2.29 MJ
(b) 119 mC
(c) 0
245
(d) 20.0 J
(c) They are the same.
(c) –114 mA
(b) 12.0 µC
(c) 37.9 mA
(d) 72.0 µ J
(b) 69.9 Ω
ε L = − LK
(b) ∆Vc =
−Kt 2
2C
(b) ~100 nH
(c) m 0Js and zero (d) m 0Js2/2
(c) t = 2 LC
(c) ~1 ns
(d) 318 Pa
33
Alternating Current Circuits
CHAPTER OUTLINE
33.1
33.2
33.3
33.4
33.5
33.6
33.7
33.8
33.9
AC Sources
Resistors in an AC Circuit
Inductors in an AC Circuit
Capacitors in an AC Circuit
The RLC Series Circuit
Power in an AC Circuit
Resonance in a Series RLC
Circuit
The Transformer and Power
Transmission
Rectifiers and Filters
ANSWERS TO QUESTIONS
∆Vmax
2
(ii) Answer (c). The average of the squared voltage is
*Q33.1 (i) Answer (d). ∆Vavg =
([ ∆V ] )
2
avg
∆Vrms
2
∆Vmax )
(
=
. Then its square root is
2
∆V
= max
2
*Q33.2 Answer (c). AC ammeters and voltmeters read rms
values. With an oscilloscope you can read a maximum
voltage, or test whether the average is zero.
*Q33.3
Q33.4
Q33.5
(i) Answer (f). The voltage varies between +170 V and −170 V.
(ii) Answer (d).
(iii) 170V/ 2 = 120 V. Answer (c).
The capacitive reactance is proportional to the inverse of the frequency. At higher and higher frequencies, the capacitive reactance approaches zero, making a capacitor behave like a wire. As the frequency
goes to zero, the capacitive reactance approaches infinity—the resistance of an open circuit.
The second letter in each word stands for the circuit element. For an inductor L, the emf ε leads
the current I—thus ELI. For a capacitor C, the current leads the voltage across the device. In a
circuit in which the capacitive reactance is larger than the inductive reactance, the current leads
the source emf—thus ICE.
Q33.6
The voltages are not added in a scalar form, but in a vector form, as shown in the phasor
diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages
across different circuit elements are not simultaneously at their maximum values. Do not forget
that an inductor can induce an emf in itself and that the voltage across it is 90° ahead of the
current in the circuit in phase.
Q33.7
In an RLC series circuit, the phase angle depends on the source frequency. At very low frequency
the capacitor dominates the impedance and the phase angle is near −90°. The phase angle is zero
at the resonance frequency, where the inductive and capacitive reactances are equal. At very high
frequencies f approaches +90°.
*Q33.8 (i) Inductive reactance doubles when frequency doubles. Answer (f).
(ii) Capacitive reactance is cut in half when frequency doubles. Answer (b).
(iii) The resistance remains unchanged. Answer (d).
*Q33.9 At resonance the inductive reactance and capacitive reactance cancel out. Answer (c).
247
248
Chapter 33
*Q33.10 At resonance the inductive reactance and capacitive reactance add to zero. tan−1(XL−XC)/R = 0.
Answer (c).
*Q33.11 (a) The circuit is in resonance. (b) 10 Ω/20 Ω = 0.5 (c) The resistance of the load could be
increased to make a greater fraction of the emf’s power go to the load. Then the emf would
put out a lot less power and less power would reach the load.
Q33.12 The person is doing work at a rate of P = Fv cos θ . One can consider the emf as the “force” that
moves the charges through the circuit, and the current as the “speed” of the moving charges. The
cos θ factor measures the effectiveness of the cause in producing the effect. Theta is an angle in
real space for the vacuum cleaner and phi is the analogous angle of phase difference between the
emf and the current in the circuit.
*Q33.13 The resonance is high-Q, so at 1 000 Hz both XL and XC are equal and much larger than R. Now XC
at 500 Hz is twice as large as at 1 kHz. And XL at 1.5 kHz is 1.5 times larger than at 1 kHz. Again,
XC at 1 500 Hz is two-thirds as large as at 1 kHz. And XL at 500 Hz is half as large as at 1 kHz. The
resistance does not change with frequency. The ranking is then a > f > b = e > c > d > g = h = i.
Q33.14 In 1881, an assassin shot President James Garfield. The bullet was lost in his body. Alexander
Graham Bell invented the metal detector in an effort to save the President’s life. The coil is
preserved in the Smithsonian Institution. The detector was thrown off by metal springs in
Garfield’s mattress, a new invention itself. Surgeons went hunting for the bullet in the wrong
place. Garfield died.
Q33.15 No. A voltage is only induced in the secondary coil if the flux through the core changes in time.
Q33.16 The Q factor determines the selectivity of the radio receiver. For example, a receiver with a
very low Q factor will respond to a wide range of frequencies and might pick up several adjacent
radio stations at the same time. To discriminate between 102.5 MHz and 102.7 MHz requires a
high-Q circuit. Typically, lowering the resistance in the circuit is the way to get a higher quality
resonance.
*Q33.17 In its intended use, the transformer takes in energy by electric transmission at 12 kV and puts out
nearly the same energy by electric transmission at 120 V. With the small generator putting energy
into the secondary side of the transformer at 120 V, the primary side has 12 kV induced across it.
It is deadly dangerous for the repairman.
SOLUTIONS TO PROBLEMS
Section 33.1
AC Sources
Section 33.2
Resistors in an AC Circuit
P33.1
∆v (t ) = ∆Vmax sin (ω t ) = 2 ∆Vrms sin (ω t ) = 200 2 sin ⎡⎣2 π (100t )⎤⎦ =
P33.2
∆Vrms =
170 V
= 120 V
2
(a)
P=
(b)
R=
( ∆Vrms )2 → R = (120 V )2 =
R
(120 V )2
100 W
75.0 W
= 144 Ω
193 Ω
(283 V) sin (628t )
Alternating Current Circuits
P33.3
Each meter reads the rms value.
100 V
= 70.7 V
2
∆Vrms 70.7 V
= 2.95 A
=
=
R
24.0 Ω
∆Vrms =
I rms
FIG. P33.3
P33.4
(a)
∆v R = ∆Vmax sin ω t
∆v R = 0.250 ( ∆Vmax ) , so, sin ω t = 0.250, or ω t = sin −1 ( 0.250 ) .
The smallest angle for which this is true is ω t = 0.253 rad. Thus, if t = 0.010 0 s,
0.253 rad
ω=
= 25.3 rad s .
0.010 0 s
(b)
The second time when ∆v R = 0.250 ( ∆Vmax ) , ω t = sin −1 ( 0.250 ) again. For this
occurrence, ω t = π − 0.253 rad = 2.89 rad (to understand why this is true, recall the
identity sin (π − θ ) = sin θ from trigonometry). Thus,
t=
P33.5
2.89 rad
= 0.114 s
25.3 rad s
iR = I max sin ω t
becomes
( 0.007 00 ) ω = sin −1 ( 0.600 ) = 0.644
Thus,
and ω = 91.9 rad s = 2 π f
P33.6
0.600 = sin (ω 0.007 00 )
so
f = 14.6 Hz
∆Vmax = 15.0 V and Rtotal = 8.20 Ω + 10.4 Ω = 18.6 Ω
I max =
∆Vmax 15.0 V
=
= 0.806 A = 2 I rms
Rtotal 18.6 Ω
⎛
2
Pspeaker = I rms
Rspeaker = ⎜
2
0.806 A ⎞
⎟ (10.4 Ω ) = 3.38 W
⎝
2 ⎠
Section 33.3
P33.7
(a)
(b)
Inductors in an AC Circuit
∆Vmax 100
=
= 13.3 Ω
7.50
I max
13.3
X
L= L =
= 0.042 4 H = 42.4 mH
ω
2π ( 50.0 )
∆Vmax 100
XL =
=
= 40.0 Ω
2.50
I max
XL =
ω=
XL
40.0
=
= 942 rad s
L
42.4 × 10 −3
249
250
P33.8
Chapter 33
⎛ X L 60.0 Hz ⎞ 50.0
At 50.0 Hz, X L = 2π ( 50.0 Hz ) L = 2π ( 50.0 Hz ) ⎜
( 54.0 Ω ) = 45.0 Ω
⎟=
⎝ 2π ( 60.0 Hz ) ⎠ 60.0
I max =
P33.9
∆Vmax
=
XL
iL ( t ) =
2 ( ∆Vrms )
=
XL
2 (100 V )
= 3.14 A
45.0 Ω
(80.0 V ) sin ⎣⎡( 65.0π ) ( 0.015 5) − π 2 ⎤⎦
∆Vmax
π
sin ⎛ ω t − ⎞ =
⎝
ωL
2⎠
( 65.0π rad s ) ( 70.0 × 10 −3 H )
iL ( t ) = ( 5.60 A ) sin (1.59 rad ) = 5.60 A
P33.10
ω = 2 π f = 2 π ( 60.0 s) = 377 rad s
X L = ω L = ( 377 s) ( 0.020 0 V ⋅ s A) = 7.54 Ω
I rms =
∆Vrms 120 V
=
= 15.9 A
7.54 Ω
XL
I max = 2 I rms = 2 (15.9 A) = 22.5 A
⎛ 2 π ( 60.0 ) 1 s ⎞
⋅
i (t ) = I max sin ω t = ( 22.5 A) sin ⎜
⎟ = ( 22.5 A) sin 120° = 19.5 A
180 ⎠
s
⎝
1
1
2
U = Li 2 = ( 0.020 0 V ⋅ s A )(19.5 A) = 3.80 J
2
2
P33.11
L=
NΦB
X ( ∆VL,max )
where Φ B is the flux through each turn. N Φ B, max = LI max = L
I
ω
XL
N Φ B,max =
Section 33.4
P33.12
P33.13
(
2 ∆VL,rms
2π f
)=
120 V ⋅ s ⎛ T ⋅ C ⋅ m ⎞ ⎛ N ⋅ m ⎞ ⎛ J ⎞
= 0.450 T ⋅ m 2
2π ( 60.0 ) ⎝ N ⋅ s ⎠ ⎝ J ⎠ ⎝ V ⋅ C ⎠
Capacitors in an AC Circuit
1
1
< 175 Ω
:
2 π f C 2 π f ( 22.0 × 10 −6 )
1
f > 41.3 Hz
< f
2π ( 22.0 × 10 −6 ) (175)
(a)
XC =
(b)
XC ∝
1
1
, so X ( 44 ) = X ( 22 ): XC < 87.5 Ω
2
C
I max = 2 I rms =
2 ( ∆Vrms )
XC
= 2 ( ∆Vrms ) 2 π f C
(a)
I max = 2 (120 V) 2 π ( 60.0 s) ( 2.20 × 10 −6 C V) = 141 mA
(b)
I max = 2 ( 240 V) 2 π ( 50.0 s) ( 2.20 × 10 −6 F ) = 235 mA
P33.14
Qmax = C ( ∆Vmax ) = C ⎡⎣ 2 ( ∆Vrms )⎤⎦ =
P33.15
I max = ( ∆Vmax ) ωC = ( 48.0 V)( 2 π ) ( 90.0 s−1 ) ( 3.70 × 10 −6 F ) = 100 mA
2C ( ∆Vrms )
Alternating Current Circuits
P33.16
XC =
1
1
=
= 2.65 Ω
ω C 2 π ( 60.0 s) (1.00 × 10 −3 C V)
∆vC ( t ) = ∆Vmax sin ω t , to be zero at t = 0
iC =
∆Vmax
sin (ω t + φ ) =
XC
2 (120 V ) ⎡
60 s −1
⎤
sin ⎢ 2π
+ 90.0° ⎥ = ( 64.0 A ) sin (120° + 90.0° )
−1
2.65 Ω
⎣ 180 s
⎦
= −32.0 A
Section 33.5 The RLC Series Circuit
P33.17
X L = ω L = 2π ( 50.0 ) ( 400 × 10 −3 ) = 126 Ω
(a)
XC =
1
1
=
= 719 Ω
ω C 2π ( 50.0 ) ( 4.43 × 10 −6 )
Z = R 2 + ( X L − XC ) = 500 2 + (126 − 719 ) = 776 Ω
2
2
∆Vmax = I max Z = ( 250 × 10 −3 ) ( 776 ) = 194 V
φ = tan −1 ⎛
⎝
(b)
X L − XC ⎞
126 − 719 ⎞
= tan −1 ⎛
= −49.9° . Thus, the
⎠
⎝
R
500 ⎠
current leads the voltage .
P33.18
ωL=
f=
P33.19
1
→ω =
ωC
1
=
LC
1
(57.0 × 10
−6
) (57.0 × 10 −6 )
= 1.75 × 10 4 rad s
ω
= 2.79 kHz
2π
(a)
X L = ω L = 2π ( 50.0 s −1 ) ( 250 × 10 −3 H ) = 78.5 Ω
(b)
XC =
(c)
Z = R 2 + ( X L − XC ) = 1.52 kΩ
(d)
I max =
(e)
φ = tan −1 ⎡⎢
⎣
−1
1
= ⎡ 2π ( 50.0 s −1 ) ( 2.00 × 10 −6 F ) ⎤⎦ = 1.59 kΩ
ωC ⎣
2
∆Vmax
210 V
=
= 138 mA
Z
1.52 × 10 3 Ω
X L − XC ⎤
= tan −1 ( −10.1) = −84.3°
R ⎥⎦
FIG. P33.17
251
252
P33.20
Chapter 33
(a)
Z = R 2 + ( X L − XC ) = 68.0 2 + (16.0 − 101) = 109 Ω
2
2
X L = ω L = (100 ) ( 0.160 ) = 16.0 Ω
1
1
XC =
=
= 101 Ω
ω C (100 ) ( 99.0 × 10 −6 )
(b)
(c)
∆Vmax 40.0 V
=
= 0.367 A
Z
109 Ω
X − XC 16.0 − 101
tan φ = L
=
= −1.25:
68.0
R
φ = −0.896 rad = −51.3°
I max =
ω = 100 rad s
I max = 0.367 A
P33.21
φ = −0.896 rad = −51.3°
X L = 2 π f L = 2 π ( 60.0 )( 0.460 ) = 173 Ω
XC =
1
1
=
= 126 Ω
2 π f C 2 π ( 60.0 ) ( 21.0 × 10 −6 )
(a)
tan φ =
X L − XC 173 Ω − 126 Ω
=
= 0.314
150 Ω
R
φ = 0.304 rad = 17.4°
(b)
P33.22
Since X L > XC , f is positive; so voltage leads the current .
For the source-capacitor circuit, the rms source voltage is ∆Vs = ( 25.1 mA ) XC . For the circuit
with resistor, ∆Vs = (15.7 mA ) R 2 + XC2 = ( 25.1 mA ) XC . This gives R = 1.247 XC . For the circuit with ideal inductor, ∆Vs = ( 68.2 mA ) X L − XC = ( 25.1 mA ) XC . So X L − XC = 0.368 0 XC .
Now for the full circuit
∆Vs = I R 2 + ( X L − XC )
2
( 25.1 mA ) XC = I (1.247 XC ) + ( 0.368 XC )
2
2
I = 19.3 mA
P33.23
XC =
1
1
=
= 1.33 × 10 8 Ω
2 π f C 2 π ( 60.0 Hz ) ( 20.0 × 10 −12 F )
Z=
(50.0 × 10
I rms =
∆Vrms
5 000 V
=
= 3.77 × 10 −5 A
1.33 × 108 Ω
Z
3
Ω ) + (1.33 × 108 Ω ) ≈ 1.33 × 108 Ω
2
2
( ∆Vrms )body = I rms Rbody = (3.77 × 10 −5 A ) (50.0 × 103 Ω ) =
1.88 V
Alternating Current Circuits
P33.24
XC =
253
1
1
=
= 49.0 Ω
ω C 2π ( 50.0 ) ( 65.0 × 10 −6 )
X L = ω L = 2π ( 50.0 ) (185 × 10 −3 ) = 58.1 Ω
Z = R 2 + ( X L − XC ) = ( 40.0 ) + ( 58.1 − 49.0 ) = 41.0 Ω
2
I max =
P33.25
2
2
∆Vmax 150
=
= 3.66 A
Z
41.0
FIG. P33.24
(a)
∆VR = I max R = ( 3.66 ) ( 40 ) = 146 V
(b)
∆VL = I max X L = ( 3.66 ) ( 58.1) = 212.5 = 212 V
(c)
∆VC = I max XC = ( 3.66 ) ( 49.0 ) = 179.1 V = 179 V
(d)
∆VL − ∆VC = 212.5 − 179.1 = 33.4 V
R = 300 Ω
500 −1 ⎞
X L = ω L = 2π ⎛
s ( 0.200 H ) = 200 Ω
⎝ π
⎠
−1
XC =
1
500 −1 ⎞
⎡
⎤
= ⎢ 2π ⎛
s (11.0 × 10 −6 F ) ⎥ = 90.9 Ω
⎝
⎠
ωC ⎣
π
⎦
Z = R + ( X L − XC ) = 319 Ω
2
2
XL = 200 Ω
{
f
R = 300 Ω
and
X − XC ⎞
φ = tan ⎛ L
= 20.0°
⎝
R ⎠
Z
XL – XC = 109 Ω
XC = 90.9 Ω
−1
FIG. P33.25
*P33.26 Let Xc represent the initial capacitive reactance. Moving the plates to half their original separation
1
doubles the capacitance and cuts XC =
in half. For the current to double, the total impedance
ωC
2
X
2
must be cut in half: Z i = 2 Z f , R 2 + ( X L − XC ) = 2 R 2 + ⎛ X L − C ⎞ . With XL = R, algebra
⎝
2 ⎠
then gives
2
⎛
⎞
X
2
R 2 + ( R − XC ) = 4 ⎜ R 2 + ⎛ R − C ⎞ ⎟
⎝
⎠
2 ⎠
⎝
2 R 2 − 2 RXC + XC2 = 8 R 2 − 4 RXC + XC2
XC = 3 R
254
Chapter 33
Section 33.6
P33.27
Power in an AC Circuit
ω = 1 000 rad s,
R = 400 Ω,
C = 5.00 × 10 −6 F, L = 0.500 H
∆Vmax = 100 V,
ω L = 500 Ω,
⎛ 1 ⎞
⎟ = 200 Ω
⎜
⎝ωC ⎠
⎛
1 ⎞
2
2
Z = R 2 + ⎜ω L −
⎟ = 400 + 300 = 500 Ω
ω
C
⎝
⎠
2
I max =
∆Vmaxx 100
=
= 0.200 A
500
Z
⎛ I2 ⎞
2
R = ⎜ max ⎟ R.
The average power dissipated in the circuit is P = I rms
⎝ 2 ⎠
( 0.200 A )2
P=
( 400 Ω ) = 8.00 W
2
P33.28
Z = R 2 + ( X L − XC ) or ( X L − XC ) = Z 2 − R 2
2
( X L − XC ) = ( 75.0 Ω )2 − ( 45.0 Ω )2 = 60.0 Ω
φ = tan −1 ⎛
⎝
I rms =
60.0 Ω ⎞
X L − XC ⎞
= tan −1 ⎛
= 53.1°
⎠
⎝
45.0 Ω ⎠
R
∆Vrms 210 V
=
= 2.80 A
75.0 Ω
Z
P = ( ∆Vrms ) I rms cos φ = ( 210 V ) ( 2.80 A ) cos ( 53.1°) = 353 W
P33.29
(a)
P = I rms ( ∆Vrms ) cos φ = ( 9.00 )180 cos ( −37.0°) = 1.29 × 103 W
2
P = I rms
R
(b)
P33.30
tan φ =
X L − XC
R
so
1.29 × 10 3 = ( 9.00 ) R
becomes
tan ( −37.0° ) =
2
and
X L − XC
: so
16
R = 16.0 Ω
X L − XC = −12.0 Ω
X L = ω L = 2π ( 60.0 s ) ( 0.025 0 H ) = 9.42 Ω
Z = R 2 + ( X L − XC ) = ( 20.0 ) + ( 9.42 ) Ω = 22.1 Ω
2
2
2
∆Vrms 120 V
=
= 5.43 A
22.1 Ω
Z
(a)
I rms =
(b)
9.42 ⎞
φ = tan −1 ⎛
= 25.2°
⎝ 20.0 ⎠
(c)
We require φ = 0. Thus, X L = XC :
9.42 Ω =
and
C = 281 µ F
(d)
Pb = Pd or ( ∆Vrms )b ( I rms )b cos φb =
( ∆Vrms )d =
so
power factor = cos φ = 0.905
1
2π ( 60.0 s −1 ) C
( ∆Vrms )d2
R
R ( ∆Vrms )b ( I rms )b cos φb = ( 20.0 Ω ) (120 V ) ( 5.43 A ) ( 0.905) = 109 V
Alternating Current Circuits
*P33.31 Consider a two-wire transmission line taking in power P
I rms =
P
∆Vrms
2
Rline =
. Then power loss = I rms
⎛ P ⎞
Thus, ⎜
⎝ ∆Vrms ⎟⎠
R1 =
2
( 2 R1 ) =
ρ ( ∆Vrms )
=
200P
A
P
100
or
R1 =
2
R1
.
∆Vrms
(∆Vrms )
200P
R1
or
A=
π ( 2r )
200 ρP =
4
( ∆Vrms )2
FIG. P33.31
800 ρP π ( ∆V )2
2r =
800(1.7 × 10 −8 Ωm) 20 000 W (18 000 m )
= 39.5 V ⋅ m/∆V
π ( ∆V )2
(a)
2r =
(b)
The diameter is inversely proportional to the potential difference.
(c)
2r = 39.5 V⋅m/1 500 V = 2.63 cm
(d)
∆V = 39.5 V⋅m/0.003 m = 13.2 kV
*P33.32 (a)
RL
2
2
and the diameter is
XL = ωL = 2p (60/s) 0.1 H = 37.7 Ω
Z = (1002 + 37.72)1/2 = 107 Ω
power factor = cosf = 100/107 = 0.936
(b)
The power factor cannot in practice be made 1.00. If the inductor were removed or if the
generator were replaced with a battery, so that either L = 0 or f = 0, the power factor would
be 1, but we would not have a magnetic buzzer.
(c)
We want resonance, with f = 0. We insert a capacitor in series with
XL = XC
P33.33
P
100
255
so
37.7 Ω = 1 s/2p C 60
C = 70.4 mF
and
One-half the time, the left side of the generator is positive, the top
diode conducts, and the bottom diode switches off. The power supply
sees resistance
⎡ 1 + 1 ⎤ = R and the power is ( ∆Vrms ) .
⎢⎣ 2 R 2 R ⎥⎦
R
R
2R
R
R
2
−1
~
The other half of the time the right side of the generator is positive,
the upper diode is an open circuit, and the lower diode has zero
resistance. The equivalent resistance is then
−1
1 1⎤
7R
Req = R + ⎡⎢
+ ⎥ =
4
⎣ 3R R ⎦
and
P=
FIG. P33.33
( ∆Vrms )2 = 4 ( ∆Vrms )2
Req
7R
2
⎡( ∆Vrms ) R ⎤ + ⎡ 4 ( ∆Vrms )2 7 R ⎤
⎣
⎦ ⎣
⎦ = 11( ∆Vrms )
2
14R
R
2
The overall time average power is:
∆V
256
Chapter 33
Section 33.7
P33.34
(a)
Resonance in a Series RLC Circuit
f=
1
2π LC
1
1
A⎛ C ⎞
C=
=
= 6.33 × 10 −13 F
2
2 2
−12
2
10
4π f L 4π (10 s ) 400 × 10 Vs ⎝ As ⎠
κ ∈0 A κ ∈0 2
=
d
d
(b)
C=
(c)
⎛ Cd ⎞
⎛ 6.33 × 10 −13 F × 10 −3 mm ⎞
−3
=
=⎜
⎜⎝
⎟⎠ = 8.46 × 10 m
1 × 8.85 × 10 −12 F
⎝ κ ∈0 ⎟⎠
X L = 2 π f L = 2 π × 1010 s × 400 × 10 −12 Vs A = 25.1 Ω
12
P33.35
ω 0 = 2π ( 99.7 × 10 6 ) = 6.26 × 108 rad s =
C=
P33.36
1
LC
1
1
=
= 1.82 pF
ω 02 L ( 6.26 × 108 )2 (1.40 × 10 −6 )
L = 20.0 mH, C = 1.00 × 10 −7, R = 20.0 Ω, ∆Vmax = 100 V
(a)
The resonant frequency for a series RLC circuit is f =
(b)
At resonance,
(c)
(d)
P33.37
12
1
2π
∆Vmax
= 5.00 A
R
ω L
From Equation 33.38, Q = 0 = 22.4
R
1
= 3.56 kHz .
LC
I max =
∆VL , max = X L I max = ω 0 LI max = 2.24 kV
The resonance frequency is ω0 =
1
. Thus, if ω = 2ω 0 ,
LC
L
⎛ 2 ⎞
XL = ω L = ⎜
L=2
⎟
⎝ LC ⎠
C
and
XC =
L
2
Z = R 2 + ( X L − XC ) = R 2 + 2.25 ⎛ ⎞
⎝ C⎠
so
I rms =
1
LC 1 L
=
=
ωC
2C
2 C
∆Vrms
=
Z
∆Vrms
R + 2.25 ( L C )
2
and the energy delivered in one period is E = P ∆t :
( ∆Vrms )2 R
⎛ 2π ⎞ = ( ∆Vrms ) RC π LC = 4π ( ∆Vrms ) RC LC
E= 2
4 R 2 C + 9.00 L
R + 2.25 ( L C ) ⎝ ω ⎠ R 2 C + 2.25 L
2
(
2
)
With the values specified for this circuit, this gives:
4π ( 50.0 V ) (10.0 Ω ) (100 × 10 −6 F )
32
2
E=
(10.0 × 10
−3
H)
4 (10.0 Ω ) (100 × 10 −6 F ) + 9.00 (10.0 × 10 −3 H )
2
12
= 242 mJ
Alternating Current Circuits
P33.38
1
.
LC
The resonance frequency is ω 0 =
Thus, if
ω = 2ω 0 ,
L
⎛ 2 ⎞
XL = ω L = ⎜
L=2
⎝ LC ⎟⎠
C
and
XC =
L
2
Then Z = R 2 + ( X L − XC ) = R 2 + 2.25 ⎛ ⎞
⎝ C⎠
so
I rms =
257
1
LC 1 L
=
=
2C
2 C
ωC
∆Vrms
=
Z
∆Vrms
R + 2.25 ( L C )
2
and the energy delivered in one period is
( ∆Vrms )2 R
⎛ 2π ⎞ = ( ∆Vrms ) RC π LC = 4π ( ∆Vrms ) RC LC
E = P ∆t = 2
4 R 2 C + 9.00 L
R + 2.25 ( L C ) ⎝ ω ⎠ R 2 C + 2.25 L
P33.39
For the circuit of Problem 20,
For the circuit of Problem 21,
2
ω0 =
2
)
(
1
=
LC
1
(160 × 10 H ) (99.0 × 10
L ( 251 rad s ) (160 × 10 H )
=
=
−3
−6
F)
= 251 rad s
−3
Q=
ω0
R
Q=
ω0 L
1 L
1
460 × 10 −3 H
L
= 0.987
=
=
=
R
R LC R C 150 Ω 21.0 × 10 −6 F
68.0 Ω
0.591
The circuit of Problem 21 has a sharper resonance.
Section 33.8
P33.40
The Transformer and Power Transmission
1
(120 V) = 9.23 V
13
(a)
∆V2,rms =
(b)
∆V1, rms I1, rms = ∆V2, rms I 2, rms
(120 V)(0.350 A) = (9.23 V) I 2,rms
I 2, rms =
(c)
P33.41
P = 42.0 W from part (b).
( ∆Vout )max =
2 000 ⎞
N2
(170 V ) = 971 V
( ∆Vin )max = ⎛⎝
350 ⎠
N1
( ∆Vout )rms = (
P33.42
42.0 W
= 4.55 A for a transformer with no energy loss.
9.23 V
971 V )
= 687 V
2
N2
(∆V1,rms )
N1
(a)
(∆V2,rms ) =
(b)
I1,rms ( ∆V1,rms ) = I 2,rms ( ∆V2,rms )
(c)
0.950 I1,rms ( ∆V1,rms ) = I 2,rms ( ∆V2,rms )
N2 =
( 2 200 ) (80 )
= 1 600 windings
110
(1.50)(2 200) = 30.0 A
I1,rms =
110
(1.20)(2 200) = 25.3 A
I1, rms =
110 ( 0.950 )
258
P33.43
Chapter 33
(a)
R = ( 4.50 × 10 −4 Ω M ) ( 6.44 × 10 5 m ) = 290 Ω and I rms =
(c)
It is impossible to transmit so much power at such low voltage. Maximum power transfer
occurs when load resistance equals the line resistance of 290 Ω, and is
3
2 ⋅ 2 ( 290 Ω )
P33.44
(a)
5.00 × 10 6 W
= 10.0 A
5.00 × 10 5 V
Ploss = I R = (10.0 A ) ( 290 Ω ) = 29.0 kW
Ploss 2.90 × 10 4
=
= 5.80 × 10 −3
P 5.00 × 106
( 4.50 × 10 V )
Section 33.9
∆Vrms
=
2
2
rms
(b)
P
2
= 17.5 kW, far below the required 5 000 kW.
Rectifiers and Filters
Input power = 8 W
Useful output power = I ∆V = 0.3 A ( 9 V ) = 2.7 W
efficiency =
(b)
useful output 2.7 W
=
= 0.34 = 34%
total input
8W
Total input power = Total output power
8 W = 2.7 W + wasted power
wasted power = 5.3 W
86 400 s ⎞ ⎛ 1 J ⎞
⎛ $0.135
= 1.29 × 108 J ⎜
⎠
⎝
⎠
⎝ 3.6 × 10 6
1d
1 Ws
⎞
⎟ = $4.8
J⎠
(c)
E = P ∆t = 8 W ( 6 ) ( 31 d ) ⎛
⎝
(a)
⎛ 1 ⎞
The input voltage is ∆Vin = IZ = I R 2 + XC2 = I R 2 + ⎜
⎟ . The output voltage is
⎝ ωC ⎠
2
P33.45
∆Vout = IR. The gain ratio is
(b)
(c)
∆Vout
IR
=
=
2
∆Vin I R + (1 ωC )2
R
R + (1 ωC )
2
2
.
∆Vout
1
→ 0 .
→ ∞ and
∆Vin
ωC
∆Vout
R
1
→ = 1 .
As ω → ∞,
→ 0 and
∆Vin
R
ωC
As ω → 0,
1
=
2
R
R 2 + (1 ωC )
1
R2 + 2 2 = 4 R2
ω C
2
ω 2C 2 =
1
3R2
ω = 2π f =
1
3RC
f=
1
2π 3 RC
Alternating Current Circuits
P33.46
(a)
The input voltage is ∆Vin = IZ = I R 2 + XC2 = I R 2 + (1 ωC ) . The output voltage is
2
∆Vout = IXC =
∆Vout
I ωC
I
=
=
. The gain ratio is
2
2
∆
V
ωC
in
I R + (1 ω C )
As ω → 0,
(c)
1
=
2
⎛ 1 ⎞
4
R2 + ⎜
⎟ = 2 2
⎝ ωC ⎠ ω C
R + (1 ω C )
2
1 ωC
R 2 + (1 ωC )
2
2
.
When
∆Vout
=
∆Vin
ω = 2π f =
3
RC
R
R + XC2
2
.
∆Vout
= 0.500,
∆Vin
0.500 Ω
then
R 2ω 2 C 2 = 3
3
2 π RC
For this RC high-pass filter,
(a)
1 ωC
2
∆Vout
1 ωC
1
→
= 1 .
→ ∞ and R becomes negligible in comparison. Then
∆Vin
1 ωC
ωC
∆Vout
1
→ 0 .
As ω → ∞,
→ 0 and
∆Vin
ωC
(b)
f=
P33.47
( 0.500 Ω )2 + XC2
= 0.500 or XC = 0.866 Ω.
If this occurs at f = 300 Hz, the capacitance is
C=
1
1
=
2π f XC 2π ( 300 Hz ) ( 0.866 Ω )
= 6.13 × 10 −4 F = 613 µ F
(b)
259
With this capacitance and a frequency of 600 Hz,
XC =
1
= 0.433 Ω
2π ( 600 Hz ) ( 6.13 × 10 −4 F )
∆Vout
=
∆Vin
R
R +X
2
2
C
=
0.500 Ω
( 0.500 Ω )2 + ( 0.433 Ω )2
= 0.756
FIG. P33.47
260
P33.48
Chapter 33
For the filter circuit,
(a)
(b)
P33.49
∆Vout
=
∆Vin
XC
R + XC2
2
.
1
1
=
= 3.32 × 10 4 Ω
−9
2 π f C 2 π ( 600 Hz ) ( 8.00 × 10 F )
At f = 600 Hz,
XC =
and
∆Vout
3.32 × 10 4 Ω
=
≈ 1.00
2
∆Vin
( 90.0 Ω )2 + ( 3.32 × 10 4 Ω )
At f = 600 kHz,
XC =
and
∆Vout
33.2 Ω
= 0.346
=
∆Vin
( 90.0 Ω )2 + ( 33.2 Ω )2
∆vout
=
∆vin
1
1
=
= 33.2 Ω
3
2 π f C 2 π ( 600 × 10 Hz ) ( 8.00 × 10 −9 F )
R
R + ( X L − XC )
2
2
At 200 Hz:
1
(8.00 Ω )
=
2
4 (8.00 Ω )2 + [ 400π L − 1 400π C ]
At 4 000 Hz:
1
⎡
⎤
2
= 4 (8.00 Ω )
(8.00 Ω ) + ⎢8 000π L −
⎥
8 000π C ⎦
⎣
2
(a)
2
At the low frequency, X L − XC < 0. This reduces to
For the high frequency half-voltage point,
Solving Equations (1) and (2) simultaneously gives
(b)
When X L = XC ,
(c)
X L = XC requires
(d)
At 200 Hz,
so the phasor diagram is as shown:
⎛R⎞
⎛1⎞
φ = − cos−1 ⎜ ⎟ = − cos−1 ⎜ ⎟ so
⎝Z ⎠
⎝2⎠
XL – XC
XL – XC
∆vout leads ∆vin by 60.0° .
At f0 , X L = XC so
∆vout and ∆vin have a phase difference of 0°° .
∆vout R 1
= = and X L − XC > 0
∆vin Z 2
1
Thus, φ = cos −1 ⎛ ⎞ = 60.0°
⎝ 2⎠
or
continued on next page
1
= −13.9 Ω
400π C
1
8 000π L −
= +13.9 Ω
8 000π C
400π L −
[1]
[2]
C = 54.6 µ F and L = 580 µ H
∆vout ⎛ ∆vout ⎞
=
= 1.00
∆vin ⎜⎝ ∆vin ⎟⎠ max
1
1
f0 =
= 894 Hz
=
2π LC 2π ( 5.80 × 10 −4 H ) ( 5.46 × 10 −5 F )
∆vout R 1
= = and XC > X L ,
∆vin Z 2
At 4 000 Hz,
FIG. P33.49(a)
2
∆vout lags ∆vin by 60.0°
R
f
or
f
∆Vout
∆Vin
Z
Z
f
or
R
FIG. P33.49(d)
∆Vin
f
∆Vout
Alternating Current Circuits
(e)
At 200 Hz and at 4 kHz,
( ∆v
P=
) = ((1 2) ∆v
2
out,rms
R
)
0
) = ( ∆v
R
=
(1 2) ⎡⎣(1 2) ∆vin,max ⎤⎦
)
(1 2) ⎡⎣ ∆vin,max ⎤⎦
2
2
out,rms
We take Q =
in,rms
R
( ∆v
At f , P =
(f )
261
in,rms
R
2
=
R
2
R
2
=
=
(10.0 V )2
= 1.56 W
8 (8.00 Ω )
(10.0 V )2
= 6.25 W
2 (8.00 Ω )
−4
ω0 L 2 π f0 L 2 π ( 894 Hz ) ( 5.80 × 10 H )
=
=
= 0.408
8.00 Ω
R
R
Additional Problems
P33.50
The equation for ∆v(t ) during the first period (using
y = mx + b) is:
∆v (t ) =
2 ( ∆Vmax ) t
T
⎡( ∆v )2 ⎤ = 1
⎣
⎦avg T
⎡⎣( ∆v ) ⎤⎦
avg
2
∆Vrms
*P33.51 (a)
− ∆Vmax
(
∫ ⎡⎣∆v (t )⎤⎦ dt =
T
2
∆Vmax )
T
0
( ∆Vmax )2 ⎛ T ⎞ [ 2t T − 1]3
=
T
2
= ⎡⎣( ∆v ) ⎤⎦avg =
⎝ 2⎠
3
( ∆Vmax )2
3
2 T
⎤
⎡2
∫ ⎢⎣T t − 1⎥⎦ dt
0
2
t =T
=
( ∆Vmax )2 ⎡( +1)3 − ( −1)3 ⎤ = ( ∆Vmax )2
t =0
=
FIG. P33.50
6
⎣
⎦
3
∆Vmax
3
Z 2 = R 2 + (XL – XC)2 760 2 = 400 2 + (700 − XC)2 417 600 = (700 − XC)2
There are two values for the square root. We can have 646.2 = 700 − XC or −646.2 = 700 − XC.
XC can be 53.8 Ω or it can be 1.35 kΩ.
(b)
If we were below resonance, with inductive reactance 700 Ω and capacitive reactance
1.35 kΩ, raising the frequency would increase the power. We must be above resonance,
with inductive reactance 700 Ω and capacitive reactance 53.8 Ω.
(c)
760 2 = 2002 + (700 − XC)2
537 600 = (700 − XC)2
Here +733 = 700 − XC has no solution
so we must have −733.2 = 700 − XC and XC = 1.43 kΩ .
262
P33.52
Chapter 33
The angular frequency is ω = 2π 60 s = 377 s. When S is open, R, L, and C are in series with the
source:
2
2
20 V ⎞
∆V
2
(1)
R 2 + ( X L − XC ) = ⎛ s ⎞ = ⎛
= 1.194 × 10 4 Ω 2
⎝ I ⎠
⎝ 0.183 A ⎠
R
When S is in position 1, a parallel combination of two R’s presents equivalent resistance , in
2
series with L and C:
2
2
⎛ R ⎞ + X − X 2 = ⎛ 20 V ⎞ = 4.504 × 10 3 Ω 2
( L C) ⎝
⎝ 2⎠
0.298 A ⎠
(2)
When S is in position 2, the current by passes the inductor. R and C are in series with the source:
2
20 V ⎞
R 2 + XC2 = ⎛
= 2.131 × 10 4 Ω 2
⎝ 0.137 A ⎠
(3)
Take equation (1) minus equation (2):
3 2
R = 7.440 × 10 3 Ω 2
R = 99.6 Ω
4
Only the positive root is physical. We have shown than only one resistance value is possible.
Now equation (3) gives
1
2 12
Only the positive root is physical and only
XC = ⎡⎣2.131 × 10 4 − ( 99.6 ) ⎤⎦ Ω = 106.7 Ω =
ωC
one capacitance is possible.
C = (ω XC ) = ⎡⎣( 377 s )106.7 Ω ⎤⎦ = 2.49 × 10 −5 F = C
−1
−1
Now equation (1) gives
2
X L − XC = ± ⎡⎣1.194 × 10 4 − ( 99.6 ) ⎤⎦ Ω = ±44.99 Ω
X L = 106.7 Ω + 44.99 Ω = 61.74 Ω or 151.7 Ω = ω L
X
L = L = 0.164 H or 0.402 H = L
ω
Two values for self-inductance are possible.
12
P33.53
ω0 =
1
=
LC
1
( 0.050 0 H ) ( 5.00 × 10 −6 F )
= 2 000 s −1
so the operating angular frequency of the circuit is
ω
ω = 0 = 1 000 s −1
2
( ∆V )2 Rω 2
Using Equation 33.37, P = 2 2 rms2 2
2
R ω + L (ω − ω 02 )
( 400 )2 (8.00 ) (1 000 )
2 = 56.7 W
2
2
(8.00 )2 (1 000 ) + ( 0.050 0 ) ⎡⎣(1.00 − 4.00 ) × 10 6 ⎤⎦
2
P=
FIG. P33.53
Alternating Current Circuits
*P33.54 (a)
At the resonance frequency XL and XC are equal. The certain frequency must be higher
than the resonance frequency for the inductive reactance to be the greater.
(b)
It is possible to determine the values for L and C, because we have three independent
equations in the three unknowns L, C, and the unknown angular frequency w.
The equations are
2 0002 = 1/LC 12 = w L and
8 = 1/w C
(c)
We eliminate w = 12/L
to have
Then 4 000 000 = 1/96 C 2 so
263
8 w C = 1 = 8(12/L)C = 96C/L so L = 96C
C = 51.0 mF and L = 4.90 mH
*P33.55 The lowest-frequency standing-wave state is NAN. The distance between the clamps we
λ
T
= f 2d.
represent as d = dNN = . The speed of transverse waves on the string is v = f λ =
2
µ
The magnetic force on the wire oscillates at 60 Hz, so the wire will oscillate in resonance at 60 Hz.
T
2
= ( 60 s ) 4 d 2
0.019 kg m
T = ( 274 kg m ⋅ s 2 ) d 2
Any values of T and d related according to this expression will work, including
if d = 0.200 m T = 10.9 N . We did not need to use the value of the current and magnetic
field. If we assume the subsection of wire in the field is 2 cm wide, we can find the rms value of
the magnetic force:
FB = I B sin θ = ( 9 A ) ( 0.02 m ) ( 0.015 3T ) sin 90° = 2.75 mN
So a small force can produce an oscillation of noticeable amplitude if internal friction is small.
f
ω L − 1 / ωC ⎞
*P33.56 φ = tan −1 ⎛
changes from −90° for w = 0 to 0
⎝
⎠
R
90°
at the resonance frequency to +90° as w goes to infinity.
The slope of the graph is df/dw:
dφ
1
1⎛
1
1 ⎞
=
⎜ L − (−1) 2 ⎟ At resonance
2
dω
ω ⎠
C
⎛ ω L − 1/ωC ⎞ R ⎝
1+ ⎜
⎟
−90°
⎝
⎠
R
we have w0 L = 1/w0 C and LC = 1/w02.
w0
FIG. P33.56
Substituting, the slope at the resonance point is
dφ
1 1⎛
1
2 L 2Q
=
=
L + LC ⎞ =
⎠ R ω0
dω ω 0 1 + 0 2 R ⎝
C
P33.57
(a)
(b)
1
will be
ωC
45.0 V
negligible compared to 200 Ω and
= 225 mA flows in the power supply and the
200 Ω
top branch.
When ω L is very large, the bottom branch carries negligible current. Also,
Now
1
→ ∞ and ω L → 0 so the generator and bottom branch carry 450 mA .
ωC
w
264
P33.58
Chapter 33
(a)
With both switches closed, the current goes only through
generator and resistor.
i (t ) =
1 ( ∆Vmax )
=
2
R
(b)
P
(c)
i (t ) =
(d)
∆Vmax
cos ω t
R
2
ω L⎞ ⎤
⎡
cos ⎢ω t + tan −1 ⎛
2
2 2
⎝
R ⎠ ⎥⎦
⎣
R +ω L
FIG. P33.58
∆Vmax
⎛ ω L − (1 ω 0 C ) ⎞
0 = φ = tan −1 ⎜ 0
⎟⎠
R
⎝
For
We require ω 0 L =
1
, so
ω0C
(e)
At this resonance frequency,
(f )
1
1
2
U = C ( ∆VC ) = CI 2 XC2
2
2
C=
1
ω 02 L
Z= R
U max
1 2 2 1 ( ∆Vmax )
1
= CI max
XC = C
=
2
2 2
2
2
ω0 C
R
(g)
U max
1 2
1 ( ∆Vmax )
= LI max
=
L
2
2
R2
(h)
Now ω = 2ω0 =
2
( ∆Vmax )2 L
2 R2
2
2
.
LC
⎛ 2 L C − (1 2 ) L C ⎞
⎛ 3 L⎞
⎛ ω L − (1 ω C ) ⎞
So φ = tan −1 ⎜
.
= tan −1 ⎜
= tan −1 ⎜
⎟
⎟
⎝
⎠
R
R
⎝ 2R C ⎟⎠
⎝
⎠
P33.59
1 1
2 ωC
ω
1
= 0
2 LC
2
(i)
Now ω L =
(a)
I R, rms =
(b)
The total current will lag the applied voltage as seen in the phasor IL
diagram at the right.
ω=
∆Vrms 100 V
=
= 1.25 A
R
80.0 Ω
∆Vrms
100 V
=
= 1.33 A
XL
2π ( 60.0 s−1 ) ( 0.200 H )
⎛I
⎞
⎛ 1.33 A ⎞
Thus, the phase angle is: φ = tan −1 ⎜⎜ L , rms ⎟⎟ = tan −1 ⎜
⎟ = 46.7° .
⎝ 1.25 A ⎠
⎝ I R,rms ⎠
I L ,rms =
IR
f
I
FIG. P33.59
∆V
Alternating Current Circuits
P33.60
Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day,
or $36 per 30 days at 10c| per kWh.) Suppose the transmission line is at 20 kV. Then
I rms =
P
∆Vrms
=
( 20 000 ) ( 500 W )
20 000 V
~10 3 A
If the transmission line had been at 200 kV, the current would be only ~10 2 A .
P33.61
P33.62
R = 200 Ω, L = 663 mH, C = 26.5 µ F, ω = 377 s −1 , ∆Vmax = 50.0 V
⎛ 1 ⎞
2
2
ω L = 250 Ω, ⎜
⎟ = 100 Ω, Z = R + ( X L − XC ) = 250 Ω
ω
C
⎝
⎠
∆Vmax 50.0 V
=
= 0.200 A
(a) I max =
250 Ω
Z
X − XC ⎞
φ = tan −1 ⎛ L
= 36.8° (∆V leads I)
⎝
R ⎠
(b)
∆VR,max = I max R = 40.0 V at φ = 0°
(c)
∆VC, max =
(d)
∆VL, max = I maxω L = 50.0 V at φ = +90.0° (∆V leads I )
I max
= 20.0 V at φ = −90.0° (I leads ∆V)
ωC
L = 2.00 H, C = 10.0 × 10 −6 F, R = 10.0 Ω, ∆v (t ) = (100 sin ω t )
(a)
The resonant frequency ω 0 produces the maximum current and thus the maximum power
delivery to the resistor.
1
1
ω0 =
= 224 rad s
=
LC
( 2.00 ) (10.0 × 10 −6 )
(b)
P=
(c)
I rms =
( ∆Vmax )2 = (100 )2
2R
∆Vrms
=
Z
2
I rms
R=
2 (10.0 )
= 500 W
∆Vrms
R + (ω L − (1 ω C ))
2
2
1 2
( I rms )max R
2
∆Vrms
R
and
( I rms )max =
or
( ∆Vrms )2 R = 1 ( ∆Vrms )2 R
Z2
2
R2
2
⎛
1 ⎞
= 2 R2
R + ⎜ω L −
⎝
ω C ⎟⎠
This occurs where Z = 2 R :
2
2
2
ω 4 L2 C 2 − 2 Lω 2 C − R 2ω 2 C 2 + 1 = 0
or
L2 C 2ω 4 − ( 2 LC + R 2 C 2 )ω 2 + 1 = 0
⎡( 2.00 )2 (10.0 × 10 −6 )2 ⎤ ω 4 − ⎡ 2 ( 2.00 ) (10.0 × 10 −6 ) + (10.0 )2 (10.0 × 10 −6 )2 ⎤ ω 2 + 1 = 0
⎣
⎦
⎣
⎦
Solving this quadratic equation, we find that
ω 2 = 51130, or 48 894
ω1 = 48 894 = 221 rad s
ω 2 = 51130 = 226 rad s
and
265
266
P33.63
Chapter 33
(a)
From Equation 33.41,
N1 ∆V1
=
.
N 2 ∆V2
Let input impedance
Z1 =
so that
N1 Z1 I1
=
N 2 Z2 I 2
∆V1
I1
and the output impedance Z 2 =
But from Eq. 33.42,
N1
=
N2
So, combining with the previous result we have
N1
=
N2
(b)
P33.64
⎛
2
P = I rms
R=
⎝
∆V2
I2
I1 ∆V2 N 2
=
=
I 2 ∆V1 N1
Z1
.
Z2
Z1
8 000
=
= 31.6
Z2
8.00
2
∆Vrms ⎞
(120 V)
R, so 250 W =
⎠
Z
Z2
2
⎛
1 ⎞
Z = R2 + ⎜ ω L −
⎝
ω C ⎟⎠
( 40.0 Ω) :
(120) ( 40.0)
2
2
( 40.0) + ⎡⎣2π f (0.185) − ⎡⎣1 2π f (65.0 × 10−6 )⎤⎦ ⎤⎦
2
2
250 =
and
250 =
1=
2 304 f 2
1 600 f + 1.3511 f 4 − 5 692.3 f 2 + 5 995 300
f =
2
2
6 396.3 ±
so
576 000 f 2
1 600 f 2 + (1.162 4 f 2 − 2 448.5)
1.3511 f 4 − 6 396.3 f 2 + 5 995 300 = 0
( 6 396.3)2 − 4 (1.3511) ( 5 995 300 )
= 3 446.5 or 1 287.4
2 (1.351 1)
f = 58.7 Hz or 35.9 Hz There are two answers because we could be above resonance or
below resonance.
P33.65
IR =
∆Vrms
;
R
IL =
∆Vrms
;
ωL
IC =
∆Vrms
(ω C )−1
(a)
⎛ 1 ⎞ ⎛
1 ⎞
2
I rms = I R2 + ( I C − I L ) = ∆Vrms ⎜ 2 ⎟ + ⎜ωC −
⎟
⎝R ⎠ ⎝
ωL ⎠
(b)
tan φ =
2
⎡ 1
IC − I L
1 ⎤⎛
1 ⎞
= ∆Vrms ⎢
−
⎥
⎜
⎟
IR
⎣ XC X L ⎦ ⎝ ∆Vrms R ⎠
⎡ 1
1 ⎤
tan φ = R ⎢
−
⎥
⎣ XC X L ⎦
FIG. P33.65
2
Alternating Current Circuits
267
*P33.66 An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor,
a capacitor, and a power supply with rms voltage 200 V and frequency
100 Hz, carries rms current 4.00 A. Find the capacitance of the capacitor.
We solve for C
2500 = 352 + (129 − 1/628C)2
1275 = (129 − 1/628C)2
There are two possibilities: 35.7 = 129 − 1/628C
1/628C = 93.1 or 1/628C = 164.5
C = either 17.1 m F or 9.67 mF
P33.67
(a)
and −35.7 = 129 − 1/628C
X L = XC = 1 884 Ω when f = 2 000 Hz
L=
XL
1 884 Ω
=
= 0.150 H and
2 π f 4 000π rad s
1
1
=
= 42.2 nF
(2π f ) XC ( 4 000π rad s)(1 884 Ω)
1
XC =
X L = 2 π f ( 0.150 H )
π
.
× 10 −8 F )
2
f
4
22
( )(
C=
Z = ( 40.0 Ω ) + ( X L − XC )
2
(b)
2
Impedance, Ω
FIG. P33.67(b)
FIG. P33.66
f (Hz) X L (Ω) XC (Ω)
300
283 12 600
600
565
6 280
800
754
4 710
1 000
942
3 770
1 500
1 410
2 510
2 000
1 880
1 880
3 000
2 830
1 260
4 000
3 770
9442
6 000
5 650
628
10 000
9 420
377
Z (Ω)
1 2300
5 720
3 960
2 830
1100
40
1 570
2 830
5 0200
9 040
268
P33.68
Chapter 33
ω0 =
1
= 1.00 × 10 6 rad s
LC
For each angular frequency, we find
Z = R 2 + (ω L − 1 ωC )
then I =
and
2
1.00 V
Z
P = I 2 (1.00 Ω )
The full width at half maximum is:
∆ω (1.000 5 − 0.999 5)ω 0
=
2π
2π
3 −1
1.00 × 10 s
∆f =
= 159 Hz
2π
∆f =
while
R
1.00 Ω
=
= 159 Hz
2π L 2π (1.00 × 10 −3 H )
ω
1
ω L (Ω)
( Ω ) Z ( Ω ) P = I 2 R ( W)
ω0
ωC
0.9990
999.0
1001.0
2.24
0.19984
0.9991
999.1
1000.9
2.06
0.235699
0.9993
999..3
1000.7
1.72
0.33768
0.9995
999.5
1000.55
1.41
0.49987
0.9997
999.7
1000.3
1.17
0.73524
0.9999
999.9
1000.1
1.022
0.96153
1.0000 1000
1000.0
1.00
1.00000
1.0001 1000.1
999.9
1.02
0.961544
0.73535
1.0003 1000.3
999.7
1.17
0.50012
1..0005 10000.5
999.5
1.41
0.33799
1.0007 1000.7
999.3
1.72
0.23601
1.0009 1000.9
999.1
2.06
0.20016
1.0010 1001
999.0
2.24
1.0
0.8
I 2 R 0.6
(W) 0.4
0.2
0.0
0.996 0.998
1
w/w 0
1.002 1.004
FIG. P33.68
*P33.69 (a)
A B
A B
We can use sin A + sin B = 2 sin ⎛ + ⎞ cos ⎛ − ⎞ to find the sum of the two sine
⎝ 2 2⎠
⎝ 2 2⎠
functions to be
E1 + E2 = ( 24.0 cm ) sin ( 4.5t + 35.0° ) cos 35.0°
E1 + E2 = (19.7 cm ) sin ( 4.5t + 35.0° )
Thus, the total wave has amplitude 19.7 cm and has a constant phase difference of
35.0° from the first wave.
(b)
In units of cm, the resultant phasor is
y R = y1 + y 2 = 12.0 ˆi + 12.0 cos 70.0° ˆi + 12.0 sin 70.0 ˆj
= 16.1ˆi + 11.3ˆj
(
) (
)
wt
y
yR
α
y2
⎛ 11.3 ⎞
2
2
y R = (16.1) + (11.3) at tan −1 ⎜
⎟ = 19.7 cm at 35.0°
⎝ 16.1 ⎠
y1
The answers are identical.
FIG. P33.69(b)
continued on next page
70.0°
x
Alternating Current Circuits
(c)
y R = 12.0 cos 70.0°ˆi + 12.0 sin 70.0°ˆj
+15.5 cos 80.0°ˆi − 15.5 sin 80.0°ˆj
+17.0 cos 160°ˆi + 17.0 sin 160° ĵ
y R = −9.18 ˆi + 1.83ˆj = 9.36 cm at 169°
The wave function of the total wave is
yR = ( 9.36 cm ) sin (15 x − 4.5t + 169° ) .
y
269
kx - w t
y1
y2
yR
y3
FIG. P33.69(c)
ANSWERS TO EVEN PROBLEMS
(b) 144 Ω
P33.2
(a)193 Ω
P33.4
(a) 25.3 rad/s
P33.6
3.38 W
P33.8
3.14 A
P33.10
3.80 J
P33.12
(a) greater than 41.3 Hz
P33.14
(b) 0.114 s
(b) less than 87.5 Ω
2C ( ∆Vrms )
P33.16
–32.0 A
P33.18
2.79 kHz
P33.20
(a) 109 Ω
P33.22
19.3 mA
P33.24
(a) 146 V
P33.26
Cutting the plate separation in half doubles the capacitance and cuts in half the capacitive
reactance to XC /2. The new impedance must be half as large as the old impedance for the new
current to be doubled. For the new impedance we then have
(b) 0.367 A
(b) 212 V
(c) I max = 0.367 A, ω = 100 rad s, φ = −0.896 rad
(c) 179 V
(d) 33.4 V
(R2 + [R − XC /2]2)1 / 2 = 0.5(R2 + [R − XC]2 )1 / 2. Solving yields XC = 3R.
P33.28
353 W
P33.30
(a) 5.43 A
P33.32
(a) 0.936 (b) Not in practice. If the inductor were removed or if the generator were replaced
with a battery, so that either L = 0 or f = 0, the power factor would be 1, but we would not have
a magnetic buzzer. (c) 70.4 mF
P33.34
(a) 633 fF
(b) 0.905
(c) 281 µ F
(b) 8.46 mm (c) 25.1 Ω
(d) 109 V
x
270
Chapter 33
P33.36
(a) 3.56 kHz
(b) 5.00 A
(c) 22.4
(d) 2.24 kV
P33.38
4π ( ∆Vrms ) RC LC
4 R2C + 9 L
P33.40
(a) 9.23 V
P33.42
(a) 1 600 turns
P33.44
(a) 0.34
P33.46
(a) See the solution.
P33.48
(a) 1.00
P33.50
See the solution.
P33.52
Only one value for R and only one value for C are possible. Two values for L are possible.
R = 99.6 Ω, C = 24.9 mF, and L = 164 mH or 402 mH
P33.54
(a) Higher. At the resonance frequency XL = XC. As the frequency increases, XL goes up and
XC goes down. (b) It is. We have three independent equations in the three unknowns L, C, and
the certain f. (c) L = 4.90 mH and C = 51.0 m F
P33.56
See the solution.
P33.58
(a) i ( t ) =
2
(d) C =
(b) 4.55 A
(c) 42.0 W
(b) 30.0 A
(c) 25.3 A
(b) 5.3 W (c) $4.8
(b) 1; 0
(c)
3
2π RC
(b) 0.346
∆Vmax
cos ω t
R
1
ω 02 L
(b) P =
(e) Z = R
(f)
( ∆Vmax )2
2R
( ∆Vmax )2 L
2R
2
ω L⎞ ⎤
⎡
cos ⎢ω t + tan −1 ⎛
⎝
R ⎠ ⎥⎦
⎣
R +ω L
∆Vmax
(c) i ( t ) =
(g)
2
( ∆Vmax )2 L
2R
2
2
2
⎛ 3 L⎞
(h) tan −1 ⎜
⎝ 2R C ⎟⎠
(i)
1
2LC
P33.60
~10 3 A
P33.62
(a) 224 rad s
P33.64
The frequency could be either 58.7 Hz or 35.9 Hz. We can be either above or below resonance.
P33.66
An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor, a capacitor, and a
power supply with rms voltage 200 V and frequency 100 Hz, carries rms current 4.00 A. Find the
capacitance of the capacitor. Answer: It could be either 17.1 mF or 9.67 mF.
P33.68
See the solution.
(b) 500 W
(c) 221 rad s and 226 rad s
34
Electromagnetic Waves
CHAPTER OUTLINE
34.1
34.2
34.3
34.4
34.5
34.6
34.7
ANSWERS TO QUESTIONS
Displacement Current and the
General Form of Ampère’s Law
Maxwell’s Equations and Hertz’s
Discoveries
Plane Electromagnetic Waves
Energy Carried by
Electromagnetic Waves
Momentum and Radiation
Pressure
Production of Electromagnetic
Waves by an Antenna
The Spectrum of Electromagnetic
Waves
*Q34.1 Maxwell included a term in Ampère’s law to account
for the contributions to the magnetic field by changing
electric fields, by treating those changing electric fields
as “displacement currents.”
*Q34.2 No, they do not. Specifically, Gauss’s law in magnetism
prohibits magnetic monopoles. If magnetic monopoles
existed, then the magnetic field lines would not have to be
closed loops, but could begin or terminate on a magnetic
monopole, as they can in Gauss’s law in electrostatics.
Q34.3
Radio waves move at the speed of light. They can
travel around the curved surface of the Earth, bouncing
between the ground and the ionosphere, which has an
altitude that is small when compared to the radius of the
Earth. The distance across the lower forty-eight states is
approximately 5 000 km, requiring a transit time of
5 × 10 6 m
~ 10 −2 s . To go halfway around the Earth
3 × 108 m s
takes only 0.07 s. In other words, a speech can be heard
on the other side of the world before it is heard at the
back of a large room.
Q34.4
Energy moves. No matter moves. You could say that electric and magnetic fields move, but it is
nicer to say that the fields at one point stay at that point and oscillate. The fields vary in time, like
sports fans in the grandstand when the crowd does the wave. The fields constitute the medium for
the wave, and energy moves.
Q34.5
The changing magnetic field of the solenoid induces eddy currents in the conducting core. This
is accompanied by I 2 R conversion of electrically-transmitted energy into internal energy in the
conductor.
*Q34.6
(i) According to f = (2π )−1 ( LC )−1/ 2, to make f half as large, the capacitance should be made four
times larger. Answer (a).
(ii) Answer (b).
*Q34.7 Answer (e). Accelerating charge, changing electric field, or changing magnetic field can be the
source of a radiated electromagnetic wave.
*Q34.8 (i) Answer (c).
(ii) Answer (c).
(iii) Answer (c).
*Q34.9 (i) through (v) have the same answer (c).
271
(iv) Answer (b).
(v) Answer (b).
272
Chapter 34
Q34.10
Sound
Light
The world of sound extends to the top of
the atmosphere and stops there; sound requires
a material medium. Sound propagates by a
chain reaction of density and pressure disturbances recreating each other. Sound in
air moves at hundreds of meters per second.
Audible sound has frequencies over a range
of three decades (ten octaves) from 20 Hz to
20 kHz. Audible sound has wavelengths of
ordinary size (1.7 cm to 17 m). Sound waves
are longitudinal.
The universe of light fills the whole universe.
Light moves through materials, but faster in a
vacuum. Light propagates by a chain reaction of electric and magnetic fields recreating
each other. Light in air moves at hundreds of
millions of meters per second. Visible light
has frequencies over a range of less than one
octave, from 430 to 750 Terahertz. Visible
light has wavelengths of very small size
(400 nm to 700 nm). Light waves are
transverse.
Sound and light can both be reflected, refracted, or absorbed to produce internal energy. Both
have amplitude and frequency set by the source, speed set by the medium, and wavelength set by
both source and medium. Sound and light both exhibit the Doppler effect, standing waves, beats,
interference, diffraction, and resonance. Both can be focused to make images. Both are described
by wave functions satisfying wave equations. Both carry energy. If the source is small, their
intensities both follow an inverse-square law. Both are waves.
Q34.11
wave. The
The Poynting vector S describes the energy flow associated with an electromagnetic
direction of S is along the direction of propagation and the magnitude of S is the rate
at which
electromagnetic energy crosses a unit surface area perpendicular to the direction of S .
*Q34.12 (i) Answer (b). Electric and magnetic fields must both carry the same energy, so their amplitudes
are proportional to each other. (ii) Answer (a). The intensity is proportional to the square of the
amplitude.
*Q34.13 (i) Answer (c). Both the light intensity and the gravitational force follow inverse-square laws.
(ii) Answer (a). The smaller grain presents less face area and feels a smaller force due to light
pressure.
Q34.14 Photons carry momentum. Recalling what we learned in Chapter 9, the impulse imparted to a
particle that bounces elastically is twice that imparted to an object that sticks to a massive wall.
Similarly, the impulse, and hence the pressure exerted by a photon reflecting from a surface must
be twice that exerted by a photon that is absorbed.
Q34.15 Different stations have transmitting antennas at different locations. For best reception align your
rabbit ears perpendicular to the straight-line path from your TV to the transmitting antenna. The
transmitted signals are also polarized. The polarization direction of the wave can be changed by
reflection from surfaces—including the atmosphere—and through Kerr rotation—a change in
polarization axis when passing through an organic substance. In your home, the plane of polarization is determined by your surroundings, so antennas need to be adjusted to align with the
polarization of the wave.
Q34.16 Consider a typical metal rod antenna for a car radio. The rod detects the electric field portion of
the carrier wave. Variations in the amplitude of the incoming radio wave cause the electrons in
the rod to vibrate with amplitudes emulating those of the carrier wave. Likewise, for frequency
modulation, the variations of the frequency of the carrier wave cause constant-amplitude vibrations of the electrons in the rod but at frequencies that imitate those of the carrier.
Electromagnetic Waves
273
*Q34.17 (i) Gamma rays have the shortest wavelength. The ranking is a < g < e < f < b < c < d.
(ii) Gamma rays have the highest frequency: d < c < b < f < e < g < a.
(iii) All electromagnetic waves have the same physical nature. a = b = c = d = e = f = g.
Q34.18 The frequency of EM waves in a microwave oven, typically 2.45 GHz, is chosen to be in a
band of frequencies absorbed by water molecules. The plastic and the glass contain no water
molecules. Plastic and glass have very different absorption frequencies from water, so they
may not absorb any significant microwave energy and remain cool to the touch.
Q34.19 People of all the world’s races have skin the same color in the infrared. When you blush or
exercise or get excited, you stand out like a beacon in an infrared group picture. The brightest
portions of your face show where you radiate the most. Your nostrils and the openings of your ear
canals are bright; brighter still are just the pupils of your eyes.
Q34.20 Light bulbs and the toaster shine brightly in the infrared. Somewhat fainter are the back of the
refrigerator and the back of the television set, while the TV screen is dark. The pipes under the sink
show the same weak sheen as the walls until you turn on the faucets. Then the pipe on the right
turns very black while that on the left develops a rich glow that quickly runs up along its length.
The food on your plate shines; so does human skin, the same color for all races. Clothing is dark as
a rule, but your bottom glows like a monkey’s rump when you get up from a chair, and you leave
behind a patch of the same blush on the chair seat. Your face shows you are lit from within, like a
jack-o-lantern: your nostrils and the openings of your ear canals are bright; brighter still are just
the pupils of your eyes.
Q34.21 12.2-cm waves have a frequency of 2.46 GHz. If the Q value of the phone is low (namely if it is
cheap), and your microwave oven is not well shielded (namely, if it is also cheap), the phone can
likely pick up interference from the oven. If the phone is well constructed and has a high Q value,
then there should be no interference at all.
SOLUTIONS TO PROBLEMS
Section 34.1
*P34.1
*P34.2
Displacement Current and the General Form of Ampère’s Law
(a)
d Φ E dQ dt
I
( 0.100 A )
=
=
=
= 11.3 × 10 9 V ⋅ m s
dt
∈0
∈0 8.85 × 10 −12 C2 N ⋅ m 2
(b)
I d = ∈0
dΦ E
= I = 0.100 A
dt
dΦ E d
dQ dt
I
= ( EA ) =
=
dt
dt
∈0
∈0
(a)
dE
I
=
= 7.19 × 1011 V m ⋅ s
dt ∈0 A
(b)
∫ B ⋅ ds = ∈
0
B=
µ0
dΦ E
⎤
d ⎡ Q
so 2π rB = ∈0 µ0 ⎢
⋅ π r2 ⎥
dt
dt ⎣ ∈0 A
⎦
−2
µ0 Ir µ0 ( 0.200 ) ( 5.00 × 10 )
=
= 2.00 × 10 −7 T
2
2A
2π ( 0.100 )
274
*P34.3
Chapter 34
We use the extended form of Ampère’s law, Equation 34.7.
Since no moving charges are present, I = 0 and we have
dΦ
∫ B ⋅ d = µ0 ∈0 dt E .
In order to evaluate the integral, we make use of the symmetry
of the situation. Symmetry requires that no particular direction
from the center can be any different from any other direction.
Therefore, there must be circular symmetry about the central axis.
We know the magnetic field lines are circles about the axis.
Therefore, as we travel around such a magnetic field circle, the
magnetic field remains constant in magnitude. Setting aside until
later the determination of the direction of B, we integrate B ⋅ dl
∫
around the circle
at
R = 0.15 m
to obtain
2π RB
Differentiating the expression
Φ E = AE
we have
d Φ E ⎛ π d 2 ⎞ dE
=⎜
⎝ 4 ⎟⎠ dt
dt
Thus,
2⎞
⎛
⎜ π d ⎟ dE
B
⋅
d
=
2
π
RB
=
µ
∈
⎜
⎟
0
0
∫
⎜⎝ 4 ⎟⎠ dt
Solving for B gives
B=
Substituting numerical values,
(4π × 10
B=
FIG. P34.3
µ0 ∈0 ⎛⎜ π d 2 ⎞⎟ dE
⎜
⎟
2π R ⎜⎝ 4 ⎟⎠ dt
−7
2
H m ) (8.85 × 10 −12 F m ) ⎡⎣π ( 0.10 m ) ⎤⎦ ( 20 V m ⋅ s)
2π ( 0.15 m )( 4 )
B = 1.85 × 10 −18 T
In Figure P34.3, the direction of the increase of the electric field
is out the plane of the
paper. By the right-hand rule, this implies that the direction of B is counterclockwise.
Thus, the direction of B at P is upwards .
Electromagnetic Waves
Section 34.2
P34.4
(a)
Maxwell’s Equations and Hertz’s Discoveries
The rod creates the same electric field that it would if stationary.
We apply Gauss’s law to a cylinder of radius r = 20 cm and length :
∫ E ⋅ dA =
qinside
∈0
E ( 2π rl ) cos 0° =
E=
(b)
λ
2π ∈0 r
λl
∈0
(35 × 10 C m ) N ⋅ m ˆj =
radially outward =
2π (8.85 × 10 C ) ( 0.2 m )
−9
FIG. P34.4
2
−12
2
3.15 × 10 3 ˆj N C
The charge in motion constitutes a current of ( 35 × 10 −9 C m ) (15 × 10 6 m s ) = 0.525 A.
This current creates a magnetic field.
µ I
B= 0
2π r
(c)
=
(4π × 10
−7
T ⋅ m A ) ( 0.525 A )
2π ( 0.2 m )
kˆ = 5.25 × 10 −7 k̂ T
The Lorentz force on the electron is F = qE + qv × B
F = (−1.6 × 10 −19 C) 3.15 × 10 3 ˆj N C
+ (−1.6 × 10 −19
(
)
C) ( 240 × 10 ˆi m s)
6
⎛
N⋅s ⎞
× ⎜ 5.25 × 10 −7 kˆ
⎟
⎝
C⋅m⎠
F = 5.04 × 10 −16 − ˆj N + 2.02 × 10 −17 + ˆj N = 4.83 × 10 −16 − ˆj N
( )
*P34.5
( )
( )
F = ma = qE + qv × B
ˆi
ˆj
kˆ
e a = ⎡⎣ E + v × B ⎤⎦ where v × B = 200
0
0 = −200 ( 0.400 ) ˆj + 200 ( 0.300 ) kˆ
m
0.200 0.300 0.400
1.60 × 10 −19
⎡50.0 ˆj − 80.0 ˆj + 60.0 kˆ ⎤ = 9.58 × 10 7 ⎡ −30.0 ˆj + 60.0 kˆ ⎤
a=
⎦
⎣
⎦
1.67 × 10 −27 ⎣
a = 2.87 × 10 9 ⎡⎣ − ˆj + 2kˆ ⎤⎦ m s 2 = −2.87 × 10 9 ˆj + 5.75 × 10 9 kˆ m s 2
(
*P34.6
F = ma = qE + qv × B
so
)
ˆi
ˆj
kˆ
−e ⎡ E + v × B ⎤⎦ where v × B = 10.0 0
0 = −4.00 ˆj
a=
m⎣
0
0 0.400
−19
( −1.60 × 10 )
⎡ 2.50 ˆi + 5.00 ˆj − 4.00 ˆj ⎤ = ( −1.76 × 1011 ) ⎡ 2.50 ˆi + 1.00 ˆj ⎤
a=
−31
⎣
⎦
⎣
⎦
9.11 × 10
a=
275
( −4.39 × 10
11
)
ˆi − 1.76 × 1011 ˆj m s 2
276
Chapter 34
Section 34.3
P34.7
Plane Electromagnetic Waves
Since the light from this star travels at 3.00 × 108 m s
(a)
the last bit of light will hit the Earth in
6.44 × 1018 m
= 2.15 × 1010 s = 680 years.
3.00 × 108 m s
Therefore, it will disappear from the sky in the year 2 007 + 680 = 2.69 × 10 3 A.D.
The star is 680 light-years away.
(b)
∆t =
∆x 1.496 × 1011 m
=
= 499 s = 8.31 min
3 × 108 m s
v
(c)
∆t =
8
∆x 2 ( 3.84 × 10 m )
= 2.56 s
=
v
3 × 108 m s
(d)
∆t =
6
∆x 2π ( 6.37 × 10 m )
= 0.133 s
=
v
3 × 108 m s
(e)
∆t =
∆x 10 × 10 3 m
=
= 3.33 × 10 −5 s
v 3 × 108 m s
P34.8
v=
1
1
=
c = 0.750 c = 2.25 × 108 m s
κµ0 ∈0
1.78
P34.9
(a)
fλ = c
f ( 50.0 m ) = 3.00 × 108 m s
f = 6.00 × 10 6 Hz = 6.00 MHz
so
(b)
or
E
=c
B
or
22.0
= 3.00 × 108
Bmax
so Bmax = −73.3kˆ nT
(c)
k=
2π
2π
=
= 0.126 m −1
λ 50.0
ω = 2π f = 2π ( 6.00 × 10 6 s −1 ) = 3.77 × 10 7 rad s
and
B = Bmax cos ( kx − ω t ) = −73.3 cos ( 0.126 x − 3.77 × 10 7 t ) k̂ nT
P34.10
E
=c
B
or
220
= 3.00 × 108
B
so B = 7.33 × 10 −7 T = 733 nT
Electromagnetic Waves
P34.11
P34.12
(a)
B=
E
100 V m
=
= 3.33 × 10 −7 T = 0.333 µ T
8
c 3.00 × 10 m s
(b)
λ=
2π
2π
=
= 0.628 µ m
k 1.00 × 10 7 m −1
(c)
f=
c 3.00 × 108 m s
=
= 4.77 × 1014 Hz
λ 6.28 × 10 −7 m
E = Emax cos ( kx − ω t )
∂E
= −Emax sin ( kx − ω t )( k )
∂x
∂E
= −Emax sin ( kx − ω t ) (−ω )
∂t
∂2 E
= −Emax cos ( kx − ω t ) ( k 2 )
2
∂x
∂2 E
2
= −Emax cos ( kx − ω t ) (−ω )
2
∂t
We must show:
∂E
∂2 E
µ
=
∈
0
0
∂x 2
∂t 2
That is,
− ( k 2 ) Emax cos ( kx − ω t ) = − µ0 ∈0 ( −ω ) Emax cos ( kx − ω t )
But this is true, because
k2 ⎛ 1 ⎞
1
= 2 = µ0 ∈0
=⎜
2
⎟
⎝ fλ⎠
ω
c
2
2
The proof for the wave of magnetic field follows precisely the same steps.
P34.13
In the fundamental mode, there is a single loop in the standing wave between the plates.
Therefore, the distance between the plates is equal to half a wavelength.
λ = 2 L = 2 ( 2.00 m ) = 4.00 m
c 3.00 × 108 m s
=
= 7.50 × 10 7 Hz = 75.0 MHz
λ
4.00 m
λ
dA to A = 6 cm ± 5% =
2
λ = 12 cm ± 5%
Thus,
P34.14
f=
v = λ f = ( 0.12 m ± 5%) ( 2.45 × 10 9 s −1 ) = 2.9 × 108 m s ± 5%
277
278
Chapter 34
Section 34.4
Energy Carried by Electromagnetic Waves
P34.15
S=I=
P34.16
Sav =
Energy
I 1 000 W m 2
=u= =
= 3.33 µ J m 3
Unit Volume
c 3.00 × 108 m s
U Uc
=
= uc
At V
P
4π r
=
2
4.00 × 10 3 W
2
2 = 7.68 µ W m
4π ( 4.00 × 1 609 m )
Emax = 2 µ0 cSav = 0.076 1 V m
∆Vmax = Emax L = ( 76.1 mV m ) ( 0.650 m ) = 49.5 mV ( ampllitude ) or 35.0 mV (rms)
P34.17
r = ( 5.00 mi ) (1 609 m mi ) = 8.04 × 10 3 m
S=
P
4π r
2
=
P
*P34.18 (a)
area
250 × 10 3 W
4π (8.04 × 10 W )
=
3
2
= 307 µ W m 2
600 × 10 3 ( J s ) h ⎛ 1 d ⎞
energy
600 kWh
2
=
=
⎜
⎟ = 6.75 W m
∆t ⋅ area ( 30 d )(13 m ) ( 9.5 m ) 30 d (123.5 m 2 ) ⎝ 24 h ⎠
gal ⎞
The car uses gasoline at the rate ( 55 mi h ) ⎛
. Its rate of energy conversion is
⎝ 25 mi ⎠
(b)
⎛ 2.54 kg ⎞
⎛ 1h ⎞
⎞
= 6.83 × 10 4 W . Its powerP = 44 × 106 J kg ⎜
( 55 mi h ) ⎛⎜⎝
⎟
⎝ 1 gal ⎟⎠
25 mi ⎠ ⎜⎝ 3 600 s ⎟⎠
per-footprint-area is
(c)
P34.19
P34.20
P
area
=
gal
6.83 × 10 4 W
= 6.64 × 10 3 W m 2
2.10 m ( 4.90 m )
.
For an automobile of typical weight and power to run on sunlight, it would have to carry
a solar panel huge compared to its own size. Rather than running a conventional car, it is
much more natural to use solar energy for agriculture, forestry, lighting, space heating,
drying, water purification, water heating, and small appliances.
Power output = (power input)(efficiency).
Thus,
Power input =
and
A=
I=
P
I
=
Power out 1.00 × 10 6 W
=
= 3.333 × 10 6 W
eff
0.300
3.33 × 10 6 W
= 3.33 × 10 3 m 2
1.00 × 10 3 W m 2
2
Bmax
c
P
=
2µ0
4π r 2
⎛ P ⎞ ⎛ 2µ 0 ⎞
(10.0 × 10 3 ) (2) ( 4 π × 10−7 ) = 5.16 × 10−10 T
Bmax = ⎜
=
⎟
⎜
⎟
2
⎝ 4π r2 ⎠ ⎝ c ⎠
4 π ( 5.00 × 10 3 ) ( 3.00 × 10 8 )
Since the magnetic field of the Earth is approximately 5 × 10 −5 T, the Earth’s field is some
100 000 times stronger.
Electromagnetic Waves
P34.21
(a)
279
P = I 2 R = 150 W
A = 2π rL = 2π ( 0.900 × 10 −3 m ) ( 0.080 0 m ) = 4.52 × 10 −4 m 2
S=
(b)
P
A
= 332 kW m 2
(points radially inward)
B=
µ 0 (1.00 )
µ0 I
=
= 222 µ T
2π r 2π ( 0.900 × 10 −3 )
E=
∆V IR
150 V
=
=
= 1.88 kV m
∆ x L 0.080 0 m
S=
Note:
EB
= 332 kW m 2
µ0
2
(3 × 106 V m )
E2
⎛ J ⎞ ⎛ C ⎞ ⎛ T ⋅ C ⋅ m⎞ ⎛ N ⋅ m⎞
I = max =
2 µ0 c 2 ( 4π × 10 −7 T ⋅ m A ) ( 3 × 108 m s ) ⎝ V ⋅ C ⎠ ⎝ A ⋅ s ⎠ ⎝ N ⋅ s ⎠ ⎝ J ⎠
2
P34.22
(a)
I = 1.19 × 1010 W m 2
P34.23
2
⎛ 5 × 10 −3 m ⎞
5
W m )π ⎜
⎟⎠ = 2.34 × 10 W
⎝
2
(b)
P = IA = (1.19 × 10
(a)
E ⋅ B = 80.0 ˆi + 32.0 ˆj − 64.0 kˆ ( N C ) ⋅ 0.200 ˆi + 0.080 0 ˆj + 0.290 kˆ µ T
10
2
)
(
)
(
E ⋅ B = (16.0 + 2.56 − 18.56 ) N 2 ⋅ s C2 ⋅ m = 0
(b)
)
(
)
(
1 ⎡⎣ 80.0 ˆi + 32.0 ˆj − 64.0 kˆ N C ⎤⎦ × ⎡⎣ 0.200 ˆi + 0.080 0 ˆj + 0.290 kˆ µ T ⎤⎦
S=
E×B=
µ0
4π × 10 −7 T ⋅ m A
)
(
6.40 kˆ − 23.2 ˆj − 6.40 kˆ + 9.28 ˆi − 12.8 ˆj + 5.12 îi × 10 −6 W m 2
S=
4π × 10 −7
S=
P34.24
(11.5ˆi − 28.6ˆj) W m
2
= 30.9 W m 2 at −68.2° from the +x axis
The energy put into the water in each container by electromagnetic radiation can be written as
eP ∆t = eIA∆t where e is the percentage absorption efficiency. This energy has the same effect as
heat in raising the temperature of the water:
eIA∆ t = mc∆T = ρVc∆T
∆T =
eI 2 ∆ t eI ∆ t
=
ρ 3 c
ρc
where is the edge dimension of the container and c the specific heat of water. For the small
container,
∆T =
(10
0.7 ( 25 × 10 3 W m 2 ) 480 s
3
kg m 3 ) ( 0.06 m ) 4 186 J kg ⋅ °C
For the larger,
∆T =
= 33.4°C
0.91 ( 25 J s ⋅ m 2 ) 480 s
( 0.12 m ) 4 186
2
J °C
= 21.7°C
280
P34.25
P34.26
Chapter 34
Emax
:
c
7.00 × 10 5 N C
= 2.33 mT
3.00 × 108 m s
(a)
Bmax =
(b)
I=
(c)
I=
(a)
E = cB = ( 3.00 × 108 m s ) (1.80 × 10 −6 T ) = 540 V m
(b)
−6
B 2 (1.80 × 10 )
= 2.58 µ J m 3
uav =
=
µ0
4π × 10 −7
(c)
Sav = cuav = ( 3.00 × 108 ) ( 2.58 × 10 −6 ) = 773 W m 2
Bmax =
( 7.00 × 10 )
=
2 ( 4π × 10 ) ( 3.00 × 10 )
5 2
2
Emax
:
2 µ0 c
I=
−7
8
650 MW m 2
⎡π
P:
⎤
P = IA = ( 6.50 × 10 8 W m 2 ) ⎢ (1.00 × 10−3 m ) ⎥ = 510 W
⎦
⎣4
A
2
2
*P34.27 (a)
We assume that the starlight moves through space without any of it being absorbed. The
radial distance is
20 ly = 20 c (1 yr ) = 20 ( 3 × 10 8 m s) ( 3.156 × 10 7 s) = 1.889 × 1017 m
I=
(b)
P
4π r2
=
4 × 10 28 W
4π (1.89 × 1017 m )
2
= 8.88 × 10 −8 W m 2
The Earth presents the projected target area of a flat circle:
P = IA = ( 8.88 × 10−8 W m 2 ) π ( 6.37 × 10 6 m ) = 1.13 × 10 7 W
2
Section 34.5
*P34.28 (a)
Momentum and Radiation Pressure
The radiation pressure is
2 (1 370 W m 2 )
3.00 × 108 m s 2
= 9.13 × 10 −6 N m 2 .
Multiplying by the total area, A = 6.00 × 10 5 m 2 gives: F = 5.48 N .
The acceleration is:
(c)
It will arrive at time t where d =
or
P34.29
a=
(b)
For complete absorption, P =
t=
F
5.48 N
=
= 9.13 × 10 −4 m s 2
m 6 000 kg
1 2
at
2
2d
=
a
2 ( 3.84 × 10 8 m )
(9.13 × 10−4 m s2 )
25.0
S
=
= 83.3 nPa .
c 3.00 × 108
= 9.17 × 10 5 s = 10.6 days
Electromagnetic Waves
*P34.30 (a)
The magnitude of the momentum transferred to the assumed totally reflecting surface in
time t is p = 2TER /c = 2SAt/c. Then the vector momentum is
p = 2SAt /c = 2(6 ˆi W/m 2 )( 40 × 10 −4 m 2 )(1 s)/(3 ×10 8 m/s)
= 1.60 × 10 −10 î kg ⋅ m/s each second
(b)
The pressure on the assumed totally reflecting surface is P = 2S/c. Then the force is
PAî = 2SAî/c = 2(6 W/m2)(40 × 10–4 m2)(1 s)/(3 × 108 m/s) = 1.60 × 10 −10 ˆi N
(c)
P34.31
I=
The answers are the same. Force is the time rate of momentum transfer.
P
πr2
=
2
Emax
2µ 0 c
P ( 2µ 0 c )
(a)
Emax =
(b)
15 × 10 −3 J s
(1.00 m ) = 50.0 pJ
3.00 × 108 m s
(c)
p=
*P34.32 (a)
πr2
= 1.90 kN C
5 × 10 −11
U
=
= 1.67 × 10 −19 kg ⋅ m s
c 3.00 × 108
The light pressure on the absorbing Earth is P =
The force is F = PA =
from the Sun.
(b)
S I
= .
c c
(1370 W/m 2 )π (6.37 × 10 6 m)2
I
π R2 ) =
= 5.82 × 108 N away
(
3.00 × 108 m s
c
The attractive gravitational force exerted on Earth by the Sun is
Fg =
−11
2
2
30
24
GM S M M ( 6.67 × 10 N ⋅ m kg ) (1.991 × 10 kg ) ( 5.98 × 10 kg )
=
rM2
(1.496 × 1011 m )2
= 3.55 × 10 22 N
which is 6.10 × 1013 times stronger and in the oppositee direction compared to the
repulsive force in part (a).
281
282
Chapter 34
*P34.33 (a)
If PS is the total power radiated by the Sun, and rE and rM are the radii of the orbits of the
planets Earth and Mars, then the intensities of the solar radiation at these planets are:
IE =
PS
4 π rE2
PS
and
IM =
Thus,
⎛r ⎞
⎛ 1.496 × 1011 m ⎞
I M = I E ⎜ E ⎟ = (1 370 W m 2 ) ⎜
= 590 W m 2
⎝ 2.28 × 1011 m ⎟⎠
⎝ rM ⎠
4 π rM2
2
(b)
2
Mars intercepts the power falling on its circular face:
PM = I M (π RM2 ) = ( 590 W m 2 ) ⎡⎣π ( 3.37 × 106 m ) ⎤⎦ = 2.10 × 1016 W
2
(c)
If Mars behaves as a perfect absorber, it feels pressure P =
and force
(d)
F = PA =
SM I M
=
c
c
P
2.10 × 1016 W
IM
π RM2 ) = M =
= 7.01 × 10 7 N
(
3.00 × 108 m s
c
c
The attractive gravitational force exerted on Mars by the Sun is
Fg =
−11
N ⋅ m 2 kg2 ) (1.991 × 10 30 kg ) ( 6.42 × 10 23 kg )
GM S M M ( 6.67 × 10
=
rM2
( 2.28 × 1011 m )2
= 1.64 × 10 21 N
which is ~1013 times stronger than the repulsive force of part (c).
(e)
P34.34
The relationship between the gravitational force and the light-pressure force is similar at
very different distances because both forces follow inverse-square laws. The force ratios are
not identical for the two planets because of their different radii and densities.
The radiation pressure on the disk is P =
πr I
.
c
Take torques about the hinge: ∑ τ = 0
2
S I F
F
= = = 2.
c c A πr
Hx
Hy
Then F =
H x ( 0 ) + H y ( 0 ) − mgr sin θ +
PA
mg
π r 2 Ir
=0
c
r
⎛ 1 kg m 2 ⎞
π r2I
π ( 0.4 m )2 10 7 W s 2 s
θ = sin
= sin −1
mgc
( 0.024 kg) m 2 ( 9.8 m ) ( 3 × 108 m ) ⎜⎝ 1 W s3 ⎟⎠
−1
q
= sin −1 0.0712 = 4.09°
FIG. P34.34
Section 34.6
P34.35
Production of Electromagnetic Waves by an Antenna
λ=
c
= 536 m
f
so
h=
λ
= 134 m
4
λ=
c
= 188 m
f
so
h=
λ
= 46.9 m
4
Electromagnetic Waves
P34.36
P=
( ∆V )2
or
R
P ∝ ( ∆V )2
∆V = ( − ) Ey ⋅ ∆y = Ey ⋅ cos θ
∆V ∝ cos θ
P34.37
so
∆y
receiving
antenna
q
P ∝ cos θ
2
(a)
θ = 15.0°: P = Pmax cos 2 (15.0° ) = 0.933Pmax = 93.3%
(b)
θ = 45.0°: P = Pmax cos 2 ( 45.0°) = 0.500Pmax = 50.0%
(c)
θ = 90.0°: P = Pmax cos 2 ( 90.0° ) = 0
(a)
Constructive interference occurs when
d cos θ = nλ for some integer n.
cos θ = n
283
FIG. P34.36
⎛ λ ⎞
λ
= n⎜
⎟ = 2n
d
⎝λ 2⎠
n = 0, ±1, ± 2, …
∴ strong signal @ θ = cos −1 0 = 90 , 270 (b)
Destructive interference occurs when
2 n + 1⎞
λ:
cos θ = 2n + 1
2 ⎠
∴ weak signal @ θ = cos −1 ( ±1) = 0 , 180 d cos θ = ⎛
⎝
FIG. P34.37
P34.38
For the proton, ΣF = ma yields
qv B sin 90.0 =
The period of the proton’s circular motion is therefore:
T=
The frequency of the proton’s motion is
The charge will radiate electromagnetic waves at this frequency, with
P34.39
(a)
f =
λ=
mv 2
R
2π R 2π m
=
v
qB
1
T
c
2π mc
= cT =
f
qB
1
The magnetic field B = µ0 J max cos ( kx − ω t ) kˆ applies for x > 0, since it describes a
2
1 E × B as
wave moving in the î direction. The electric field direction must satisfy S =
µ0
ˆi = ˆj × kˆ so the direction of the electric field is ĵ when the cosine is positive. For its
1
magnitude we have E = cB , so altogether we have E = µ0 cJ max cos ( kx − ω t ) ˆj .
2
continued on next page
284
Chapter 34
(b)
1 1 1 2 2
S=
E×B=
µ cJ cos 2 ( kx − ω t ) ˆi
µ0
µ0 4 0 max
1
2
S = µ0 cJ max
cos 2 ( kx − ω t ) ˆi
4
(c)
The intensity is the magnitude of the Poynting vector averaged over one or more cycles.
1
1
2
The average of the cosine-squared function is , so I = µ0 cJ max
.
8
2
(d)
Section 34.7
P34.40
J max =
8I
=
µ0 c
8 ( 570 W m 2 )
4π × 10 −7 ( Tm A ) 3 × 108 m s
= 3.48 A m
The Spectrum of Electromagnetic Waves
From the electromagnetic spectrum chart and accompanying text discussion, the following
identifications are made:
c
f
Frequency, f
Wavelength, λ =
2 Hz = 2 × 100 Hz
150 Mm
Radio
2 kHz = 2 × 103 Hz
150 km
Radio
2 MHz = 2 × 106 Hz
150 m
Radio
2 GHz = 2 × 109 Hz
15 cm
Microwave
2 THz = 2 × 1012 Hz
150 mm
Infrared
2 PHz = 2 × 1015 Hz
150 nm
Ultraviolet
2 EHz = 2 × 1018 Hz
150 pm
X-ray
2 ZHz = 2 × 1021 Hz
150 fm
Gamma ray
2 YHz = 2 × 1024 Hz
150 am
Gamma ray
Wavelengh, l
Frequency, f =
2 km = 2 × 103 m
1.5 × 105 Hz
Radio
2 m = 2 × 100 m
1.5 × 108 Hz
Radio
2 mm = 2 × 10−3 m
1.5 × 1011 Hz
Microwave
2 mm = 2 × 10−6 m
1.5 × 1014 Hz
Infrared
2 nm = 2 × 10−9 m
1.5 × 1017 Hz
Ultraviolet or X-ray
2 pm = 2 × 10−12 m
1.5 × 1020 Hz
X-ray or Gamma ray
2 fm = 2 × 10−15 m
1.5 × 1023 Hz
Gamma ray
2 am = 2 × 10−18 m
1.5 × 1026 Hz
Gamma ray
c
λ
Classification
Classification
Electromagnetic Waves
P34.41
(a)
fλ = c
gives
(5.00 × 10
19
285
Hz ) λ = 3.00 × 108 m s :
λ = 6.00 × 10 −12 m = 6.00 pm
(b)
fλ = c
gives
(4.00 × 10
9
Hz ) λ = 3.00 × 108 m s:
λ = 0.075 m = 7.50 cm
P34.42
f =
(b)
1 000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 × 10 −5 m thick.
f =
*P34.43 (a)
(b)
(c)
P34.44
c 3 × 108 m s
=
~108 Hz
λ
1.7 m
(a)
3.00 × 108 m s
~ 1013 Hz
6 × 10 −5 m
Channel 4:
Channel 6:
Channel 8:
radio wave
infrared
fmin = 66 MHz
λmax = 4.55 m
fmax = 72 MHz
λmin = 4.17 m
fmin = 82 MHz
λmax = 3.66 m
fmax = 88 MHz
λmin = 3.41 m
fmin = 180 MHz
λmax = 1.67 m
fmax = 186 MHz
λmin = 1.61 m
The time for the radio signal to travel 100 km is:
∆t r =
100 × 10 3 m
= 3.33 × 10 −4 s
3.00 × 108 m s
The sound wave travels 3.00 m across the room in:
∆t s =
3.00 m
= 8.75 × 10 −3 s
343 m s
Therefore, listeners 100 km away will receive the news before the people in the newsroom by a
total time difference of ∆t = 8.75 × 10 −3 s − 3.33 × 10 −4 s = 8.41 × 10 −3 s.
P34.45
The wavelength of an ELF wave of frequency 75.0 Hz is λ =
c 3.00 × 108 m s
=
= 4.00 × 10 6 m.
f
75.0 Hz
The length of a quarter-wavelength antenna would be
L = 1.00 × 10 6 m = 1.00 × 10 3 km
or
⎛ 0.621 mi ⎞
L = (1 000 km ) ⎜
⎟ = 621 mi
⎝ 1.00 km ⎠
Thus, while the project may be theoretically possible, it is not very practical.
286
Chapter 34
Additional Problems
P34.46
ω = 2π f = 6.00π × 10 9 s −1 = 1.88 × 1010 s −1
k=
2π ω 6.00π × 10 9 s −1
= =
= 20.0π = 62.8 m −1
λ
c 3.00 × 108 m s
E = ( 300 V m ) cos ( 62.8 x − 1.88 × 1010 t )
*P34.47 (a)
(b)
Bmax =
E
300 V m
=
= 1.00 µ T
c 3.00 × 108 m s
B = (1.00 µ T ) cos ( 62.8 x − 1.88 × 1010 t )
P = (1 370 W m 2 ) ⎡⎣ 4π (1.496 × 1011 m ) ⎤⎦ = 3.85 × 10 26 W
2
P = SA:
2 ( 4π × 10 −7 N A 2 ) (1 370 W m 2 )
2 µ0 S
=
c
S=
2
cBmax
2 µ0
so
Bmax =
S=
2
Emax
2µ0 c
so
Emax = 2 µ0 cS = 2 ( 4π × 10 −7 ) ( 3.00 × 108 ) (1 370 ) = 1.02 kV m
3.00 × 108 m s
= 3.39 µ T
*P34.48 Suppose you cover a 1.7 m × 0.3 m section of beach blanket. Suppose the elevation angle of the
Sun is 60°. Then the target area you fill in the Sun’s field of view is
(1.7 m )(0.3 m ) cos 30 = 0.4 m 2
Now I =
P
A
=
U
At
U = IAt = (1 370 W m 2 ) ⎡⎣( 0.6 )( 0.5 ) ( 0.4 m 2 )⎤⎦ ( 3 600 s) ~10 6 J
y
*P34.49 (a)
(d) u
x
0
l
2l
(a) Ey
(b)
u E = 12 ∈0 E 2 =
(c)
uB =
(d)
2
u = u E + u B = ∈0 Emax
cos 2 ( kx )
(e)
Eλ =
1
2
2
∈0 Emax
cos 2 ( kx )
2
µ ∈ 2
1 2
1 2
1 Emax
B =
Bmax cos 2 ( kx ) =
cos 2 ( kx ) = 0 0 Emax
cos 2 ( kx ) = u E
2µ0
2µ0
2µ0 c 2
2µ0
∫
λ
0
2
∈ 0 Emax
cos 2 ( kx ) Adx =
λ
2
Ax0 +
= 12 ∈ 0 Emax
=
(f)
I=
1
2
∫
λ
0
2
∈ 0 Emax
[ 12 + 12 cos((2 kx )] Adx
2
A
∈ 0 Emax
∈ E2 A
λ
2
sin(2 kx ) 0 = 12 ∈ 0 Emax
Aλ + 0 max [sin( 4 π ) − sin(0 )]
4k
4k
2
∈ 0 Emax
Aλ
2
Eλ ∈0 Emax
Aλ
=
=
2 AT
AT
1
2
2
∈0 cEmax
This result agrees with equation 34.24, I =
2
Emax
cE 2
cE 2 µ ∈
= max2 = max 0 0 .
2µ 0 c 2µ 0 c
2µ 0
Electromagnetic Waves
P34.50
(a)
Fgrav =
287
GM S m ⎛ GM S ⎞ ⎡ ⎛ 4 3 ⎞ ⎤
=
ρ πr ⎥
⎝ R 2 ⎠ ⎢⎣ ⎝ 3
⎠⎦
R2
where M S = mass of Sun, r = radius of particle, and R = distance from Sun to particle.
Frad =
Since
Sπ r 2
,
c
Frad ⎛ 1 ⎞ ⎛ 3SR 2 ⎞ 1
=
∝
Fgrav ⎝ r ⎠ ⎜⎝ 4 cGM S ρ ⎟⎠ r
(b)
From the result found in part (a), when Fgrav = Frad ,
we have r =
r=
3SR 2
4 cGM S ρ
3 ( 214 W m 2 ) ( 3.75 × 1011 m )
2
4 ( 6.67 × 10 −11 N ⋅ m 2 kg2 ) (1.991 × 10 30 kg ) (1 500 kg m 3 ) ( 3.00 × 108 m s )
= 3.78 × 10 −7 m
P34.51
Emax
= 6.67 × 10 −16 T
c
(a)
Bmax =
(b)
Sav =
(c)
P = Sav A = 1.67 × 10−14 W
(d)
F = PA =
2
Emax
= 5.31 × 10 −17 W m 2
2 µ0 c
⎛ Sav ⎞
A = 5.56 × 10 −23 N (approximately the
⎝ c ⎠
weight of 3 000 hydrogen atoms!)
*P34.52 (a)
FIG. P34.51
In E = q/4p ∈0r 2 , the net flux is q/∈0, so
2
2
E = Φ r̂ /4p r 2 = (487 N ⋅ m2/C) r̂ /4p r 2 = (38.8/r ) rˆ N ⋅ m /C
(b)
The radiated intensity is I = P /4p r 2 = E 2max/2m0c. Then
Emax = (P m 0c /2p )1/2/r
= [(25 N ⋅ m/s)(4p × 10–7 T ⋅ m/2p A)(3 × 108 m/s)(1 N ⋅ s/1 T ⋅ C ⋅ m)(1 A ⋅ s/1 C)]1/2/r
= (38.7/r ) N ⋅ m/C
(c)
For 3 × 106 N/C = (38.7 N ⋅ m/C)/r we find r = 12.9 µ m , but the expression in part (b)
does not apply if this point is inside the source.
(d)
In the radiated wave, the field amplitude is inversely proportional to distance. As the distance
doubles, the amplitude is cut in half and the intensity is reduced by a factor of 4. In the static
case, the field is inversely proportional to the square of distance. As the distance doubles, the
field is reduced by a factor of 4. The intensity of radiated energy is everywhere zero.
288
Chapter 34
2u
= 95.1 mV m
∈0
P34.53
2
u = 12 ∈0 Emax
P34.54
The area over which we model the antenna as radiating is the lateral surface of a cylinder,
Emax =
A = 2 π r = 2 π ( 4.00 × 10 −2 m ) ( 0.100 m ) = 2.51 × 10 −2 m 2
(a)
The intensity is then: S =
(b)
The standard is
P
A
=
0.600 W
= 23.9 W m 2 .
2.51 × 10 −2 m 2
⎛ 1.00 × 10 −3 W ⎞ ⎛ 1.00 × 10 4 cm 2 ⎞
2
0.570 mW cm 2 = 0.570 ( mW cm 2 ) ⎜
⎟⎠ = 5.70 W m
⎝ 1.00 mW ⎟⎠ ⎜⎝
1.00 m 2
23.9 W m 2
= 4.19 times .
5.70 W m 2
While it is on, the telephone is over the standard by
P34.55
(a)
Bmax =
Emax
175 V m
=
= 5.83 × 10 −7 T
3.00 × 108 m s
c
2π
2π
ω = kc = 1.26 × 1011 rad s
=
= 419 rad m
λ
0.015 0 m
Since S is along x, and E is along y, B must be in the z direction . (That is, S ∝ E × B. )
k=
S av =
(b)
Sav =
Emax Bmax
= 40.6 W m 2
2 µ0
(c)
Pr =
2S
= 2.71 × 10 −7 N m 2
c
(d)
a=
a=
∑ F = PA = ( 2.71 × 10
m
( 406
m
−7
N m 2 ) ( 0.750 m 2 )
0.500 kg
( 40.6
W m 2 ) ˆi
= 4.06 × 10 −7 m s 2
nm s 2 ) ˆi
S = 1 370 W m 2
*P34.56 Of the intensity
the 38.0% that is reflected exerts a pressure
P1 =
2 Sr 2 ( 0.380 ) S
=
c
c
The absorbed light exerts pressure
P2 =
Sa 0.620 S
=
c
c
Altogether the pressure at the subsolar point on Earth is
2
1.38S 1.38 (1 370 W m )
=
= 6.30 × 10 −6 Pa
8
3.00 × 10 m s
c
(a)
Ptotal = P1 + P2 =
(b)
Pa
1.01 × 10 5 N m 2
=
= 1.60 × 1010 times smaller than atmospheric pressure
Ptotal 6.30 × 10 −6 N m 2
Electromagnetic Waves
P34.57
(a)
F I
=
A c
P=
a = 3.03 × 10 −9 m s 2 and
(b)
IA P
100 J s
= =
= 3.33 × 10 −7 N = (110 kg ) a
c
c 3.00 × 108 m s
x=
1 2
at
2
0 = (107 kg ) v − ( 3.00 kg ) (12.0 m s − v ) = (107 kg ) v − 36.0 kg ⋅ m s + ( 3.00 kg ) v
v=
P34.58
F=
2x
= 8.12 × 10 4 s = 22.6 h
a
t=
36.0
= 0.327 m s
110
t = 30.6 s
The mirror intercepts power P = I1 A1 = (1.00 × 10 3 W m 2 ) ⎡⎣π ( 0.500 m )2 ⎤⎦ = 785 W .
In the image,
(a)
I2 =
(b)
I2 =
P
A2
:
2
Emax
so
2 µ0 c
I2 =
785 W
2
2 = 625 kW m
π ( 0.020 0 m )
Emax = 2 µ0 cI 2 = 2 ( 4π × 10 −7 ) ( 3.00 × 108 ) ( 6.25 × 10 5 )
= 21.7 kN C
Bmax =
(c)
0.400
Emax
= 72.4 µ T
c
P ∆t = mc∆T
0.400 ( 785 W ) ∆t = (1.00 kg ) ( 4 186 J kg ⋅ C ) (100 C − 20.0 C )
∆t =
P34.59
3.35 × 10 5 J
= 1.07 × 10 3 s = 17.8 min
314 W
Think of light going up and being absorbed by the bead which presents a face area π rb2
The light pressure is P =
289
S I
= .
c c
4 ρ gc ⎛ 3m ⎞
I=
3 ⎜⎝ 4π ρ ⎟⎠
(a)
I π rb2
4
F =
= mg = ρ π rb3 g
c
3
(b)
P = IA = (8.32 × 10 7 W m 2 ) π ( 2.00 × 10 −3 m ) = 1.05 kW
and
2
13
= 8.32 × 10 7 W m 2
290
P34.60
Chapter 34
Think of light going up and being absorbed by the bead, which presents face area π rb2 .
If we take the bead to be perfectly absorbing, the light pressure is P =
(a)
F = Fg
F c Fg c mgc
=
=
π rb2
A
A
I=
so
From the definition of density, ρ =
m
m
=
V ( 4 3) π rb3
1 ⎛ ( 4 3) π ρ ⎞
=
rb ⎜⎝
m ⎟⎠
so
(b)
P34.61
(a)
(b)
Sav I F
.
= =
c
c A
mgc ⎛ 4π ρ ⎞
π ⎝ 3m ⎠
Substituting for rb ,
I=
P = IA
P=
4ρ
= gc ⎛ ⎞
⎝ 3⎠
4π r 2 ρ gc ⎛ 3m ⎞
⎜⎝ 4πρ ⎟⎠
3
On the right side of the equation,
F = ma = qE or
23
(C
⎛ m⎞
⎝π⎠
13
=
4 ρ gc ⎛ 3m ⎞
3 ⎜⎝ 4π ρ ⎟⎠
13
13
C2 ( m s2 )
2
23
13
2
N ⋅ m 2 ) ( m s)
3
N ⋅ m 2 ⋅ C 2 ⋅ m 2 ⋅ s3 N ⋅ m J
=
= = W.
s
s
C2 ⋅ s4 ⋅ m 3
=
−19
qE (1.60 × 10 C ) (100 N C )
= 1.76 × 1013 m s 2
a=
=
9.11 × 10 −31 kg
m
1.60 × 10 −19 ) (1.76 × 1013 )
(
q2a2
=
=
3
6π ∈0 c3 6π (8.85 × 10 −12 ) ( 3.00 × 108 )
2
The radiated power is then: P
2
= 1.75 × 10 −27 W
(c)
⎛ v2 ⎞
F = mac = m ⎜ ⎟ = qv B
⎝ r ⎠
so
v=
qBr
m
−19
v 2 q 2 B 2 r (1.60 × 10 ) ( 0.350 ) ( 0.500 )
The proton accelerates at a =
=
=
r
m2
(1.67 × 10 −27 )2
2
2
= 5.62 × 1014 m s 2
(1.60 × 10 −19 ) (5.62 × 1014 ) = 1.80 × 10 −24 W
q2a2
=
3
6π ∈0 c3 6π (8.85 × 10 −12 ) ( 3.00 × 108 )
2
The proton then radiates P =
2
Electromagnetic Waves
P34.62
291
f = 90.0 MHz, Emax = 2.00 × 10 −3 V m = 200 mV m
(a)
λ=
c
= 3.33 m
f
1
= 1.11 × 10 −8 s = 11.1 ns
f
E
Bmax = max = 6.67 × 10 −12 T = 6.67 pT
c
T=
(b)
x
t ⎞ˆ
E = ( 2.00 mV m ) cos 2π ⎛
j
−
⎝ 3.33 m 11.1 ns ⎠
x
t ⎞
B = ( 6.67 pT ) kˆ cos 2π ⎛
−
⎝ 3.33 m 11.1 ns ⎠
(c)
( 2.00 × 10 )
E2
I = max =
= 5.31 × 10 −9 W m 2
2 µ0 c 2 ( 4π × 10 −7 ) ( 3.00 × 108 )
(d)
I = cuav
(e)
P=
(a)
ε = − dΦ B
−3 2
P34.63
so
uav = 1.77 × 10 −17 J m 3
−9
2 I ( 2 ) ( 5.31 × 10 )
= 3.54 × 10 −17 Pa
=
3.00 × 108
c
dt
=−
d
( BA cos θ )
dt
ε (t ) = 2π fBmax A sin 2π f t cos θ
Thus,
ε = − A d ( Bmax cos ω t cos θ ) = ABmaxω (sin ω t cos θ )
dt
ε (t ) = 2π 2 r 2 fBmax cos θ sin 2π f t
ε max = 2π 2 r 2 f Bmax cos θ
where θ is the angle between the magnetic field and the normal to the loop.
(b)
If E is vertical, B is horizontal, so the plane of the loop should be vertical
and the plane should contain the line of sight of the transmitter .
292
P34.64
Chapter 34
(a)
m = ρV = ρ
⎛ 6m ⎞
r=⎜
⎝ ρ 4π ⎟⎠
14 3
πr
23
13
⎛
6 (8.7 kg ) ⎞
=⎜
⎟
3
⎝ ( 990 kg m ) 4π ⎠
13
= 0.161 m
1
2
4π r 2 = 2π ( 0.161 m ) = 0.163 m 2
2
(b)
A=
(c)
I = eσ T 4 = 0.970 ( 5.67 × 10 −8 W m 2 ⋅ K 4 ) ( 304 K ) = 470 W m 2
(d)
P = IA = ( 470 W m 2 ) 0.163 m 2 = 76.8 W
(e)
I=
4
2
Emax
2 µ0 c
Emax = ( 2 µ0 cI ) = ⎡⎣(8π × 10 −7 Tm A ) ( 3 × 108 m s ) ( 470 W m 2 ) ⎤⎦
12
12
(f)
Emax = cBmax
Bmax =
(g)
(h)
= 595 N C
595 N C
= 1.98 µ T
3 × 108 m s
The sleeping cats are uncharged and nonmagnetic. They carry no macroscopic current.
They are a source of infrared radiation. They glow not by visible-light emission but by
infrared emission.
⎛ 6 ( 0.8 ) ⎞
Each kitten has radius rk = ⎜
⎝ 990 × 4π ⎟⎠
13
= 0.072 8 m and radiating area
⎛ 6 ( 5.5) ⎞
2
2π ( 0.072 8 m ) = 0.033 3 m 2 . Eliza has area 2π ⎜
⎝ 990 × 4π ⎟⎠
23
= 0.120 m. The
total glowing area is 0.120 m 2 + 4 ( 0.033 3 m 2 ) = 0.254 m 2 and has power output
P = IA = ( 470 W m 2 ) 0.254 m 2 = 119 W
P34.65
(a)
At steady state,
Pin = Pout and the power radiated out is Pout = eσ AT 4.
Thus,
0.900 (1 000 W m 2 ) A = 0.700 ( 5.67 × 10 −8 W m 2 ⋅ K 4 ) AT 4
or
⎡
⎤
900 W m 2
T =⎢
−8
2
4 ⎥
⎢⎣ 0.700 ( 5.67 × 10 W m ⋅ K ) ⎥⎦
14
(b)
= 388 K = 115°C
The box of horizontal area A presents projected area A sin 50.0° perpendicular to the
sunlight. Then by the same reasoning,
0.900 (1 000 W m 2 ) A sin 50.0° = 0.700 ( 5.67 × 10 −8 W m 2 ⋅ K 4 ) AT 4
⎡
(900 W m 2 ) sin 50.0° ⎤ = 363 K = 90.0° C
T =⎢
−8
2
4 ⎥
⎢⎣ 0.700 ( 5.67 × 10 W m ⋅ K ) ⎥⎦
14
or
Electromagnetic Waves
P34.66
293
We take R to be the planet’s distance from its star. The planet, of radius r, presents a
projected area π r 2 perpendicular to the starlight. It radiates over area 4π r 2 .
At steady-state,
eI in (π r 2 ) = eσ ( 4π r 2 ) T 4
Pin = Pout:
⎛ 6.00 × 10 23 W ⎞
2
2
4
23
2 4
e⎜
⎟⎠ (π r ) = eσ ( 4π r ) T so that 6.00 × 10 W = 16πσ R T
⎝
4π R 2
R=
6.00 × 10 23 W
6.00 × 10 23 W
=
= 4.77 × 10 9 m = 4.77 Gm
16πσ T 4
16π ( 5.67 × 10 −8 W m 2 ⋅ K 4 ) ( 310 K )4
ANSWERS TO EVEN PROBLEMS
P34.2
(a) 7.19 × 1011 V m ⋅ s
P34.4
(a) 3.15 ĵ kN C
P34.6
( −4.39ˆi − 1.76ˆj)10
P34.8
2.25 × 108 m/s
P34.10
733 nT
P34.12
See the solution.
P34.14
2.9 × 108 m s ± 5%
P34.16
49.5 mV
P34.18
(a) 6.75 W/m2 (b) 6.64 kW/m2 (c) A powerful automobile running on sunlight would have to
carry on its roof a solar panel huge compared to the size of the car. Agriculture and forestry for
food and fuels, space heating of large and small buildings, water heating, and heating for drying
and many other processes are clear present and potential applications of solar energy.
P34.20
516 pT, ~10 5 times stronger than the Earth’s field
P34.22
(a) 11.9 GW m 2
P34.24
33.4°C for the smaller container and 21.7°C for the larger
P34.26
(a) 540 V/m (b) 2.58 ␮ J/m3
P34.28
(a) 5.48 N away from the Sun
P34.30
(a) 1.60 × 10−10 î kg⋅m/s each second (b) 1.60 × 10−10 î N
Force is the time rate of momentum transfer.
P34.32
(a) 582 MN away from the Sun
the opposite direction.
(b) 200 nT
(b) 525 nTk̂
11
(c) −483 ĵ aN
m s2
(b) 234 kW
(c) 773 W/m2
(b) 913 µm/s2 away from the Sun
(c) 10.6 d
(c) The answers are the same.
(b) The gravitational force is 6.10 × 1013 times stronger and in
294
Chapter 34
P34.34
4.09°
P34.36
(a) 93.3%
P34.38
2π m p c
(b) 50.0%
(c) 0
eB
P34.40
radio, radio, radio, radio or microwave, infrared, ultraviolet, x-ray, γ -ray, γ -ray; radio, radio,
microwave, infrared, ultraviolet or x-ray, x- or γ -ray, γ -ray, γ -ray
P34.42
(a) ~108 Hz radio wave
P34.44
The radio audience gets the news 8.41 ms sooner.
P34.46
E = ( 300 V m ) cos ( 62.8 x − 1.88 × 1010 t )
P34.48
~10 6 J
P34.50
(a) See the solution.
P34.52
B = (1.00 µ T ) cos ( 62.8 x − 1.88 × 1010 t )
(b) 378 nm
(a) E = (38.8/r2) r̂ N ⋅ m2/C (b) Emax = (38.7/r) (W⋅T ⋅ m2/A ⋅ s)1/2 = (38.7/r) N ⋅ m/C
(c) 12.9 µm, but the expression in part (b) does not apply if this point is inside the source.
(d) In the radiated wave, the field amplitude is inversely proportional to distance. As the distance
doubles, the amplitude is cut in half and the intensity is reduced by a factor of 4. In the static
case, the field is inversely proportional to the square of distance. As the distance doubles, the field
is reduced by a factor of 4. The intensity of radiated energy is everywhere zero in the static case.
P34.54
(a) 23.9 W m 2
P34.56
(a) 6.30 µ Pa
P34.58
(a) 625 kW m 2
P34.60
(b) ~1013 Hz infrared light
⎛ 16 m ρ 2 ⎞
(a) ⎜
⎝ 9π ⎟⎠
(b) 4.19 times the standard
(b) 1.60 × 1010 times less than atmospheric pressure
13
gc
(b) 21.7 kN C and 72.4 µ T
⎛ 16π 2 m ρ 2 ⎞
(b) ⎜
⎟⎠
⎝
9
P34.62 (a) 3.33 m, 11.1 ns, 6.67 pT
(c) 17.8 min
13
r 2 gc
⎛ x
t ⎞ˆ
j;
−
(b) E = ( 2.00 mV m ) cos 2π ⎜
⎝ 3.33 m 11.1 ns ⎟⎠
x
t ⎞
−
B = ( 6.67 pT ) kˆ cos 2π ⎛
⎝ 3.33 m 11.1 ns ⎠
(c) 5.31 nW m 2
(d) 1.77 × 10 −17 J m 3
(e) 3.54 × 10 −17 Pa
P34.64
(a) 16.1 cm (b) 0.163 m 2 (c) 470 W m 2 (d) 76.8 W (e) 595 N/C (f) 1.98 µ T
(g) The cats are nonmagnetic and carry no macroscopic charge or current. Oscillating charges
within molecules make them emit infrared radiation. (h) 119 W
P34.66
p r 2; 4p r 2 where r is the radius of the planet; 4.77 Gm
35
The Nature of Light and the
Laws of Geometric Optics
CHAPTER OUTLINE
35.1
35.2
35.3
35.4
35.5
35.6
35.7
35.8
The Nature of Light
Measurements of the Speed
of Light
The Ray Approximation in
Geometric Optics
The Wave Under Reflection
The Wave Under Refraction
Huygens’s Principle
Dispersion
Total Internal Reflection
ANSWERS TO QUESTIONS
Q35.1
Light travels through a vacuum at a speed of 300 000 km
per second. Thus, an image we see from a distant star
or galaxy must have been generated some time ago. For
example, the star Altair is 16 light-years away; if we
look at an image of Altair today, we know only what
was happening 16 years ago. This may not initially seem
significant, but astronomers who look at other galaxies
can gain an idea of what galaxies looked like when they
were significantly younger. Thus, it actually makes sense
to speak of “looking backward in time.”
*Q35.2 104 m /(3 × 108 m /s) is 33 µs. Answer (c).
*Q35.3 We consider the quantity l /d. The smaller it is, the better the ray approximation works. In (a) it is
like 0.34 m/1 m ≈ 0.3. In (b) we can have 0.7 µm/2 mm ≈ 0.000 3. In (c), 0.4 µm/2 mm ≈ 0.000 2.
In (d), 300 m/1 m ≈ 300. In (e) 1 nm/1 mm ≈ 0.000 001. The ranking is then e, c, b, a, d.
Q35.4
With a vertical shop window, streetlights and his own reflection
can impede the window shopper’s clear view of the display.
The tilted shop window can put these reflections out of the way.
Windows of airport control towers are also tilted like this, as are
automobile windshields.
FIG. Q35.4
295
296
Chapter 35
Q35.5
We assume that you and the child are
always standing close together. For a
flat wall to make an echo of a sound
that you make, you must be standing along a normal to the wall. You
must be on the order of 100 m away,
to make the transit time sufficiently
long that you can hear the echo separately from the original sound. Your
(a)
sound must be loud enough so that
you can hear it even at this considerable range. In the picture, the dashed
rectangle represents an area in which
you can be standing. The arrows
represent rays of sound.
Now suppose two vertical
perpendicular walls form an inside
corner that you can see. Some of
the sound you radiate horizontally
will be headed generally toward the
corner. It will reflect from both walls
with high efficiency to reverse in
direction and come back to you. You
can stand anywhere reasonably far
away to hear a retroreflected echo of
sound you produce.
If the two walls are not perpendicular, the inside corner will not
produce retroreflection. You will
generally hear no echo of your shout
or clap.
(b)
If two perpendicular walls have
FIG. Q35.5
a reasonably narrow gap between
them at the corner, you can still hear
a clear echo. It is not the corner line itself that retroreflects the sound, but the perpendicular walls
on both sides of the corner. Diagram (b) applies also in this case.
Q35.6
The stealth fighter is designed so that adjacent panels are not joined at right angles, to prevent any
retroreflection of radar signals. This means that radar signals directed at the fighter will not be
channeled back toward the detector by reflection. Just as with sound, radar signals can be treated
as diverging rays, so that any ray that is by chance reflected back to the detector will be too weak
in intensity to distinguish from background noise. This author is still waiting for the automotive
industry to utilize this technology.
*Q35.7 Snell originally stated his law in terms of cosecants. From v = c/n and sinθ = 1/cscθ and λ = c/nf
with c and f constant between media, we conclude that a, b, and c are all correct statements.
Q35.8
An echo is an example of the reflection of sound. Hearing the noise of a distant highway on a
cold morning, when you cannot hear it after the ground warms up, is an example of acoustical
refraction. You can use a rubber inner tube inflated with helium as an acoustical lens to concentrate sound in the way a lens can focus light. At your next party, see if you can experimentally
find the approximate focal point!
The Nature of Light and the Laws of Geometric Optics
297
*Q35.9 (a) Yes. (b) No. (c) Yes. (d) No. If the light moves into a medium of higher refractive index, its
wavelength decreases. The frequency remains constant. The speed diminishes by a factor equal
to the index of refraction. If its angle of incidence is 0°, it will continue in the same direction.
Q35.10 If a laser beam enters a sugar solution with a concentration gradient (density and index of refraction increasing with depth) then the laser beam will be progressively bent downward (toward the
normal) as it passes into regions of greater index of refraction.
*Q35.11 (a) Yes. It must be traveling in the medium in which it moves slower, water, to undergo total
internal reflection.
(b) Yes. It must be traveling in the medium in which it moves slower, air, to undergo total internal
reflection.
Q35.12 Diamond has higher index of refraction than glass and consequently a smaller critical angle
for total internal reflection. A brilliant-cut diamond is shaped to admit light from above, reflect
it totally at the converging facets on the underside of the jewel, and let the light escape only at
the top. Glass will have less light internally reflected.
Q35.13 Highly silvered mirrors reflect about 98% of the incident light. With a 2-mirror periscope,
that results in approximately a 4% decrease in intensity of light as the light passes through the
periscope. This may not seem like much, but in low-light conditions, that lost light may mean
the difference between being able to distinguish an enemy armada or an iceberg from the sky
beyond. Using prisms results in total internal reflection, meaning that 100% of the incident light
is reflected through the periscope. That is the “total” in total internal reflection.
*Q35.14 The light with the greater change in speed will have the larger deviation. Since the glass has a
higher index than the surrounding air, A travels slower in the glass.
Q35.15 Immediately around the dark shadow of my head, I see a halo brighter than the rest of the dewy
grass. It is called the heiligenschein. Cellini believed that it was a miraculous sign of divine
favor pertaining to him alone. Apparently none of the people to whom he showed it told him
that they could see halos around their own shadows but not around Cellini’s. Thoreau knew that
each person had his own halo. He did not draw any ray diagrams but assumed that it was entirely
natural. Between Cellini’s time and Thoreau’s, the Enlightenment and Newton’s explanation
of the rainbow had happened. Today the effect is easy to see whenever your shadow falls on a
retroreflecting traffic sign, license plate, or road stripe. When a bicyclist’s shadow falls on a paint
stripe marking the edge of the road, her halo races along with her. It is a shame that few people
are sufficiently curious observers of the natural world to have noticed the phenomenon.
Q35.16 At the altitude of the plane the surface of the Earth need not block off the lower half of the
rainbow. Thus, the full circle can be seen. You can see such a rainbow by climbing on a stepladder above a garden sprinkler in the middle of a sunny day. Set the sprinkler for fine mist. Do not
let the slippery children fall from the ladder.
*Q35.17 Light from the lamps along the edges of the sheet enters the plastic. Then it is totally internally
reflected by the front and back faces of the plastic, wherever the plastic has an interface with air.
If the refractive index of the grease is intermediate between 1.55 and 1.00, some of this light can
leave the plastic into the grease and leave the grease into the air. The surface of the grease is
rough, so the grease can send out light in all directions. The customer sees the grease shining
against a black background. The spotlight method of producing the same effect is much less
efficient. With it, much of the light from the spotlight is absorbed by the blackboard. The
refractive index of the grease must be less than 1.55. Perhaps the best choice would be
1.55 × 1.00 = 1.24 .
298
Chapter 35
*Q35.18 Answer (c). We want a big difference between indices of refraction to have total internal reflection under the widest range of conditions.
Q35.19 A mirage occurs when light changes direction as it moves between batches of air having different
indices of refraction because they have different densities at different temperatures. When the sun
makes a blacktop road hot, an apparent wet spot is bright due to refraction of light from the bright
sky. The light, originally headed a little below the horizontal, always bends up as it first enters
and then leaves sequentially hotter, lower-density, lower-index layers of air closer to the road
surface.
SOLUTIONS TO PROBLEMS
Section 35.1
The Nature of Light
Section 35.2
Measurements of the Speed of Light
*P35.1
The Moon’s radius is 1.74 × 10 6 m and the Earth’s radius is 6.37 × 10 6 m. The total distance
traveled by the light is:
d = 2 ( 3.84 × 108 m − 1.74 × 10 6 m − 6.37 × 10 6 m ) = 7.52 × 108 m
This takes 2.51 s, so v =
7.52 × 108 m
= 2.995 × 108 m s = 299.5 Mm s . The sizes of the
2.51 s
objects need to be taken into account. Otherwise the answer would be too large by 2%.
c=
8
∆x 2 (1.50 × 10 km ) (1 000 m km )
=
= 2.27 × 108 m s = 227 Mm s
t
( 22.0 min ) ( 60.0 s min )
P35.2
∆ x = ct ;
P35.3
The experiment is most convincing if the wheel turns fast enough to pass outgoing light through
2
one notch and returning light through the next: t =
c
⎛ 2 ⎞
θ =ωt =ω⎜ ⎟
⎝ c⎠
so
(
)
2.998 × 10 8 ⎡⎣ 2π ( 720 ) ⎤⎦
cθ
ω=
=
= 114 rad s
2
2 11.45 × 10 3
(
)
The returning light would be blocked by a tooth at one-half the angular speed, giving another data
point.
The Nature of Light and the Laws of Geometric Optics
Section 35.3
The Ray Approximation in Geometric Optics
Section 35.4
The Wave Under Reflection
Section 35.5
The Wave Under Refraction
P35.4
(a)
(b)
Let AB be the originally horizontal ceiling, BC its originally vertical
normal, AD the new ceiling, and DE its normal. Then angle BAD = φ .
By definition DE is perpendicular to AD and BC is perpendicular to AB.
Then the angle between DE extended and BC is φ because angles are
equal when their sides are perpendicular, right side to right side and
left side to left side.
Now CBE = φ is the angle of incidence of the vertical light beam.
Its angle of reflection is also φ. The angle between the vertical incident beam and the reflected beam is 2φ .
299
B
A
D
E
C
FIG. P35.4 (a)
A
D
E
FIG. P35.4 (b)
P35.5
(c)
1.40 cm
tan 2φ =
= 0.001 94
720 cm
(a)
From geometry,
1.25 m = d sin 40.0°
so
d = 1.94 m
(b)
φ = 0.055 7°
Mirror 2
50.0° above the horizontal
1.25 m
or parallel to the incident ray.
50°
Light
beam
40.0°
Mirror 2
i2 = 50°
40°
P
d
40° i1 = 40°
50°
Mirror 1
1.25 m
50°
Mirror 1
FIG. P35.5
P35.6
(a)
Method One:
α = 90° − θ1
The incident ray makes angle
with the first mirror. In the picture, the law of reflection
implies that
θ1 = θ1′
FIG. P35.6
Then
β = 90° − θ1′ = 90 − θ1 = α
In the triangle made by the mirrors and the ray passing between them,
Further,
and
β + 90° + γ = 180°
γ = 90° − β
δ = 90° − γ = β = α
∈= δ = α
Thus the final ray makes the same angle with the first mirror as did the incident ray. Its
direction is opposite to the incident ray.
Method Two:
The vector velocity of the incident light has a component v y perpendicular to the first mirror
and a component v x perpendicular to the second. The v y component is reversed upon the first
continued on next page
300
Chapter 35
reflection, which leaves v x unchanged. The second reflection reverses v x and leaves v y
unchanged. The doubly reflected ray then has velocity opposite to the incident ray.
(b)
P35.7
The incident ray has velocity v x ˆi + v y ˆj + vz kˆ . Each reflection reverses one component
and leaves the other two unchanged. After all the reflections, the light has velocity
ˆ opposite to the incident ray.
− v x ˆi − v y ˆj − vz k,
Let d represent the perpendicular distance
from the person to the mirror. The distance
between lamp and person measured parallel
to the mirror can be written in two ways:
2d tan θ + d tan θ = d tan φ . The condition on the
distance traveled by the light
d tan f
is 2 d = 2 d + d . We have the two
cos φ cos θ cos θ
equations 3 tan θ = tan φ and 2 cos θ = 3 cos φ .
To eliminate φ we write
9 sin 2 θ sin 2 φ
=
cos 2 θ
cos 2 φ
f
q
2d tan q
q
d tan q
2d
d
d
FIG. P35.7
4 cos 2 θ = 9 cos 2 φ
9 cos 2 φ sin 2 θ = cos 2 θ (1 − cos 2 φ )
4
4 cos 2 θ sin 2 θ = cos 2 θ ⎛ 1 − cos 2 θ ⎞
⎝
⎠
9
4
4 sin 2 θ = 1 − (1 − sin 2 θ ) 36 sin 2 θ = 9 − 4 + 4 sin 2 θ
9
5
sin 2 θ =
θ = 23.3°
32
*P35.8
The excess time the second pulse spends in the ice is 6.20 m/[(3.00 × 108 m/s)/1.309] = 27.1 ns
P35.9
Using Snell’s law,
sin θ 2 =
n1
sin θ1
n2
θ 2 = 25.5°
λ2 =
λ1
= 442 nm
n1
FIG. P35.9
The Nature of Light and the Laws of Geometric Optics
301
*P35.10 The law of refraction n1 sin θ1 = n2 sin θ 2 can be put into the more general form
c
c
sin θ1 = sin θ 2
v1
v2
sin θ1 sin θ 2
=
v1
v2
In this form it applies to all kinds of waves that move through space.
sin 3.5°
sin θ 2
=
343 m s 1 493 m s
sin θ 2 = 0.266
θ 2 = 15.4°
The wave keeps constant frequency in
f =
v1 v2
=
λ1 λ2
λ2 =
v2 λ1 1 493 m s ( 0.589 m )
=
= 2.56 m
v1
343 m s
The light wave slows down as it moves from air into water but the sound speeds up by a large
factor. The light wave bends toward the normal and its wavelength shortens, but the sound wave
bends away from the normal and its wavelength increases.
P35.11
n1 sin θ1 = n2 sin θ 2
sin θ1 = 1.333 sin 45°
sin θ1 = (1.33)( 0.707 ) = 0.943
θ1 = 70.5° → 19.5° above the horizon
FIG. P35.11
P35.12
P35.13
(a)
c 3.00 × 10 8 m s
f = =
= 4.74 × 1014 Hz
λ 6.328 × 10 −7 m
(b)
λglass =
λair 632.8 nm
=
= 422 nm
n
1.50
(c)
vglass =
cair 3.00 × 108 m s
=
= 2.00 × 108 m s = 200 Mm s
n
1.50
We find the angle of incidence:
n1 sin θ1 = n2 sin θ 2
1.333 sin θ1 = 1.52 sin 19.6°
θ1 = 22.5°
The angle of reflection of the beam in water is then also 22.5° .
302
Chapter 35
*P35.14 (a)
As measured from the diagram, the incidence angle is 60°, and the refraction angle is 35°.
From Snell’s law, sin θ 2 = v2 , then sin 35° = v2 and the speed of light in the block
sin 60°
c
sin θ1 v1
is 2.0 × 10 8 m s .
(b)
The frequency of the light does not change upon refraction. Knowing the wavelength in a
vacuum, we can use the speed of light in a vacuum to determine the frequency: c = f λ ,
(
)
thus 3.00 × 10 8 = f 632.8 × 10 −9 , so the frequency is 474.1 THz .
(c)
To find the wavelength of light in the block, we use the same wave speed relation, v = f λ ,
(
)
so 2.0 × 10 8 = 4.741 × 1014 λ , so λglass = 420 nm .
P35.15
c 3.00 × 108 m s
=
= 1.81 × 108 m s = 181 Mm s
n
1.66
(a)
Flint Glass:
v=
(b)
Water:
v=
c 3.00 × 108 m s
=
= 2.25 × 108 m s = 225 Mm s
n
1.333
(c)
Cubic Zirconia:
v=
c 3.00 × 108 m s
=
= 1.36 × 108 m s = 136 Mm s
n
2.20
⎛n
⎞
*P35.16 From Snell’s law, sin θ = ⎜ medium ⎟ sin 50.0°
⎝ nliver ⎠
But
so
nmedium c vmedium
v
=
= liver = 0.900
nliver
c vliver
vmedium
12 cm
50°
n liver
h
θ = sin −1 ⎡⎣( 0.900 ) sin 50.0° ⎤⎦ = 43.6°
From the law of reflection,
12.0 cm
d=
= 6.00 cm, and
2
h=
P35.17
nmedium
d
q
Tumor
FIG. P35.16
d
6.00 cm
=
= 6.30 cm
tan θ tan ( 43.6° )
n1 sin θ1 = n2 sin θ 2 :
⎛ n sin θ1 ⎞
θ 2 = sin −1 ⎜ 1
⎝ n2 ⎟⎠
⎧1.00 sin 30° ⎫
θ 2 = sin −1 ⎨
⎬ = 19.5°
⎩ 1.50
⎭
θ 2 and θ 3 are alternate interior angles formed by the ray cutting
parallel normals.
So,
θ 3 = θ 2 = 19.5°
1.50 sin θ 3 = 1.00 sin θ 4
θ 4 = 30.0°
FIG. P35.17
The Nature of Light and the Laws of Geometric Optics
P35.18
sin θ1 = nw sin θ 2
q 1 = 62°
1
1
sin θ 2 =
sin θ1 =
sin ( 90.0° − 28.0° ) = 0.662
1.333
1.333
θ 2 = sin −1 ( 0.662 ) = 41.5°
h=
d
3.00 m
=
= 3.39 m
tan θ 2 tan441.5°
air
n = 1.00
28°
q2
water
n = 1.333
h
3.0 m
FIG. P35.18
P35.19
At entry,
n1 sin θ1 = n2 sin θ 2
or
1.00 sin 30.0° = 1.50 sin θ 2
θ 2 = 19.5°
The distance h the light travels in the medium is given by
cos θ 2 =
or
h=
2.00 cm
h
2.00 cm
= 2.12 cm
cos19.5°
FIG. P35.19
α = θ1 − θ 2 = 30.0° − 19.5° = 10.5°
The angle of deviation upon entry is
The offset distance comes from sin α =
P35.20
P35.21
d
:
h
d = ( 2.21 cm ) sin 10.5° = 0.388 cm
The distance h traveled by the light is
h=
2.00 cm
= 2.12 cm
cos19.5°
The speed of light in the material is
v=
c 3.00 × 108 m s
=
= 2.00 × 108 m s
n
1.50
Therefore,
t=
h
2.12 × 10 −2 m
=
= 1.06 × 10 −10 s = 106 ps
v 2.00 × 108 m s
Applying Snell’s law at the air-oil interface,
nair sin θ = noil sin 20.0°
yields
θ = 30.4°
Applying Snell’s law at the oil-water interface
nw sin θ ′ = noil sin 20.0°
yields
303
θ ′ = 22.3°
FIG. P35.21
304
P35.22
Chapter 35
For sheets 1 and 2 as described,
n1 sin 26.5° = n2 sin 31.7°
0.849 n1 = n2
For the trial with sheets 3 and 2,
n3 sin 26.5° = n2 sin 36.7°
0.747 n3 = n2
Now
0.747n3 = 0.849n1
n3 = 1.14 n1
For the third trial,
n1 sin 26.5° = n3 sin θ 3 = 1.14 n1 sin θ 3
θ 3 = 23.1°
*P35.23 Refraction proceeds according to
(a)
For the normal component of velocity to
be constant,
(1.00 ) sin θ1 = (1.66 ) sin θ 2
v1 cos θ1 = v2 cos θ 2
or
( c ) cos θ1 = ⎛⎜⎝
We multiply Equations (1) and (2), obtaining:
sin θ1 cos θ1 = sin θ 2 cos θ 2
or
sin 2θ1 = sin 2θ 2
c ⎞
⎟ cos θ 2
1.66 ⎠
The solution θ1 = θ 2 = 0 does not satisfy Equation (2) and must be rejected. The physical
solution is 2θ1 = 180° − 2θ 2 or θ 2 = 90.0° − θ1. Then Equation (1) becomes:
sin θ1 = 1.66 cos θ1, or tan θ1 = 1.66
which yields
θ1 = 58.9°
In this case, yes , the perpendicular velocity component does remain constant.
(b)
(1)
Light entering the glass slows down and makes a smaller angle with the normal. Both
effects reduce the velocity component parallel to the surface of the glass.
Then
no, the parallel velocity component cannot remain constant, or will remain constant
.
onnly in the trivial case θ1 = θ 2 = 0
(2)
The Nature of Light and the Laws of Geometric Optics
P35.24
305
Consider glass with an index of refraction of 1.5, which is 3 mm thick. The speed of light in the
glass is
3 × 108 m s
= 2 × 108 m s
1.5
3 × 10 −3 m
3 × 10 −3 m
−
~10 −11 s
2 × 108 m s 3 × 108 m s
600 nm
For light of wavelength 600 nm in vacuum and wavelength
= 400 nm in glass,
1.5
The extra travel time is
3 × 10 −3 m 3 × 10 −3 m
−
~10 3 wavelengths
4 × 10 −7 m 6 × 10 −7 m
the extra optical path, in wavelengths, is
P35.25
Taking Φ to be the apex angle and δ min to be the angle of minimum deviation, from Equation
35.9, the index of refraction of the prism material is
n=
Solving for δ min,
P35.26
sin ⎡⎣( Φ + δ min ) 2 ⎤⎦
sin ( Φ 2 )
Φ⎞
⎛
δ min = 2 sin −1 ⎜ n sin ⎟ − Φ = 2 sin −1 ⎡⎣( 2.20 ) sin ( 25.0° ) ⎤⎦ − 50.0° = 86.8°
⎝
2⎠
n ( 700 nm ) = 1.458
(a)
(b)
(1.00 ) sin 75.0° = 1.458 sin θ 2; θ 2 =
60.0°
41.5°
Let
θ 3 + β = 90.0°, θ 2 + α = 90.0° then α + β + 60.0° = 180°
So
60.0° − θ 2 − θ 3 = 0 ⇒ 60.0° − 41.5° = θ 3 = 18.5°
q1
α
β
q2
q3
FIG. P35.26
θ 4 = 27.6°
(c)
1.458 sin 18.5° = 1.00 sin θ 4
(d)
γ = (θ1 − θ 2 ) + ⎡⎣ β − ( 90.0° − θ 4 ) ⎤⎦
γ = 75.0° − 41.5° + ( 90.0° − 18.5° ) − ( 90.0° − 27.6° ) = 42.6°
P35.27
At the first refraction,
1.00 sin θ1 = n sin θ 2
The critical angle at the second surface is given by n sin θ 3 = 1.00:
or
⎛ 1.00 ⎞
= 41.8°
θ 3 = sin −1 ⎜
⎝ 1.50 ⎟⎠
But,
θ 2 = 60.0° − θ 3
FIG. P35.27
Thus, to avoid total internal reflection at the second surface (i.e., have θ 3 < 41.8°)
it is necessary that
θ 2 > 18.2°
Since sin θ1 = n sin θ 2, this becomes
sin θ1 > 1.50 sin 18.2° = 0.468
or
θ1 > 27.9°
q4
γ
306
P35.28
Chapter 35
At the first refraction,
1.00 sin θ1 = n sin θ 2
The critical angle at the second surface is given by
Φ
q1
n sin θ 3 = 1.00, or
⎛ 1.00 ⎞
θ 3 = sin −1 ⎜
⎝ n ⎟⎠
But
( 90.0° − θ 2 ) + ( 90.0° − θ 3 ) + Φ = 180°
which gives
θ2 = Φ − θ3
q2
q3
FIG. P35.28
⎛ 1.00 ⎞
Thus, to have θ 3 < sin −1 ⎜
and avoid total internal reflection at the second surface,
⎝ n ⎟⎠
⎛ 1.00 ⎞
it is necessary that
θ 2 > Φ − sin −1 ⎜
⎝ n ⎟⎠
P35.29
Since sin θ1 = n sin θ 2, this requirement becomes
⎡
⎛ 1.00 ⎞ ⎤
sin θ1 > n sin ⎢ Φ − sin −1 ⎜
⎝ n ⎟⎠ ⎥⎦
⎣
or
⎛
⎡
⎛ 1.00 ⎞ ⎤⎞
θ1 > sin −1 ⎜ n sin ⎢ Φ − sin −1 ⎜
⎝ n ⎟⎠ ⎥⎦⎟⎠
⎝
⎣
Through the application of trigonometric identities,
θ1 > sin −1
(
n 2 − 1 sin Φ − cos Φ
Note for use in every part:
Φ + ( 90.0° − θ 2 ) + ( 90.0° − θ 3 ) = 180°
so
θ3 = Φ − θ2
At the first surface the deviation is
α = θ1 − θ 2
At exit, the deviation is
β = θ4 − θ3
The total deviation is therefore
δ = α + β = θ1 + θ 4 − θ 2 − θ 3 = θ1 + θ 4 − Φ
(a)
At entry:
n1 sin θ1 = n2 sin θ 2
Thus,
θ 3 = 60.0° − 30.0° = 30.0°
At exit:
1.50 sin 30.0° = 1.00 sin θ 4
)
FIG. P35.29
or
⎛ sin 48.6° ⎞
θ 2 = sin −1 ⎜
= 30.0°
⎝ 1.50 ⎟⎠
or
θ 4 = sin −1 ⎡⎣1.50 sin ( 30.0° ) ⎤⎦ = 48.6°
so the path through the prism is symmetric when θ1 = 48.6°.
(b)
δ = 48.6° + 48.6° − 60.0° = 37.2°
(c)
At entry:
sin θ 2 =
At exit:
sin θ 4 = 1.50 sin ( 31.6° ) ⇒ θ 4 = 51.7°
continued on next page
sin 45.6°
⇒ θ 2 = 28.4°
1.50
θ 3 = 60.0° − 28.4° = 31.6°
δ = 45.6° + 51.7° − 60.0° = 37.3°
The Nature of Light and the Laws of Geometric Optics
(d)
Section 35.6
P35.30
sin 51.6°
⇒ θ 2 = 31.5°
1.50
At entry:
sin θ 2 =
At exit:
sin θ 4 = 1.50 sin ( 28.5° ) ⇒ θ 4 = 45.7°
307
θ 3 = 60.0° − 31.5° = 28.5°
δ = 51.6° + 45.7° − 60.0° = 37.3°
Huygens’s Principle
(a)
For the diagrams of contour lines and wave fronts and rays, see Figures (a) and (b) below.
As the waves move to shallower water, the wave fronts bend to become more nearly
parallel to the contour lines.
(b)
For the diagrams of contour lines and wave fronts and rays, see Figures (c) and (d) below.
We suppose that the headlands are steep underwater, as they are above water. The rays are
everywhere perpendicular to the wave fronts of the incoming refracting waves. As shown,
the rays bend toward the headlands and deliver more energy per length at the headlands.
FIG. P35.30
Section 35.7
P35.31
Dispersion
For the incoming ray,
sin θ 2 =
sin θ1
n
Using the figure to the right,
(θ 2 )violet = sin −1 ⎛⎜⎝
sin 50.0° ⎞
⎟ = 27.48°
1.66 ⎠
(θ 2 )red = sin −1 ⎛⎜⎝
sin 50.0° ⎞
⎟ = 28.22°
1.62 ⎠
For the outgoing ray,
θ 3 = 60.0° − θ 2
and sin θ 4 = n sin θ 3:
(θ 4 )violet = sin −1 [1.66 sin 32.52°] = 63.17°
FIG. P35.31
(θ 4 )red = sin −1 [1.62 sin 31.78°] = 58.56°
The angular dispersion is the difference
∆θ 4 = (θ 4 )violet − (θ 4 )red = 63.17° − 58.56° =
4.61°
308
P35.32
Chapter 35
From Fig 35.21 nv = 1.470 at 400 nm
1.00 sin θ = 1.470 sin θ v
Then
and
nr = 1.458 at 700 nm
and
1.00 sin θ = 1.458 sin θ r
sin θ ⎞
sin θ ⎞
− sin −1 ⎛
δ r − δ v = θ r − θ v = sin −1 ⎛
⎝ 1.458 ⎠
⎝ 1.4770 ⎠
sin 30.0° ⎞
sin 30.0° ⎞
− sin −1 ⎛
= 0.171°
∆δ = sin −1 ⎛
⎝ 1.458 ⎠
⎝ 1.470 ⎠
Section 35.8
P35.33
Total Internal Reflection
n sin θ = 1. From Table 35.1,
(a)
⎛ 1 ⎞
= 24.4°
θ = sin −1 ⎜
⎝ 2.419 ⎟⎠
(b)
⎛ 1 ⎞
= 37.0°
θ = sin −1 ⎜
⎝ 1.66 ⎟⎠
(c)
⎛ 1 ⎞
= 49.8°
θ = sin −1 ⎜
⎝ 1.309 ⎟⎠
n1 sin θ1 = n2 sin 90.0°
*P35.34 For total internal reflection,
1.50 sin θ1 = 1.33 (1.00 )
P35.35
sin θ c =
or
θ1 = 62.5°
n2
n1
n2 = n1 sin 88.8° = (1.000 3)( 0.999 8 ) = 1.000 08
P35.36
sin θ c =
nair 1.00
=
= 0.735
npipe 1.36
FIG. P35.35
θ c = 47.3°
Geometry shows that the angle of refraction at the end is
φ = 90.0° − θ c = 90.0° − 47.3° = 42.7°
Then, Snell’s law at the end,
gives
1.00 sin θ = 1.36 sin 42.7°
θ = 67.2°
The 2-mm diameter is unnecessary information.
FIG. P35.36
The Nature of Light and the Laws of Geometric Optics
P35.37
P35.38
(a)
A ray along the inner edge will escape if any ray escapes. Its angle of
R−d
incidence is described by sin θ =
and by n sin θ > 1 sin 90°. Then
R
n(R − d)
nd
R>
nR − R > nd
>1
nR − nd > R
n −1
R
(b)
As d → 0, Rmin → 0.
This is reasonable.
As n increases, Rmin decreases.
This is reasonable.
As n decreases toward 1, Rmin increases.
This is reasonable.
(
1.40 100 × 10 −6 m
)=
FIG. P35.37
350 × 10 −6 m
(c)
Rmin =
(a)
sin θ 2 v2
=
sin θ1 v1
and
θ 2 = 90.0° at the critical angle
sin 90.0° 1 850 m s
=
sin θ c
343 m s
so
θ c = sin −1 ( 0.185 ) = 10.7°
0.40
309
(b)
Sound can be totally reflected if it is traveling in the medium where it travels slower: air .
(c)
Sound in air falling on the wall from most directions is 100% reflected , so the wall is a
good mirror.
P35.39
For plastic with index of refraction n ≥ 1.42 surrounded by air, the critical angle for total internal reflection is given by
1
1 ⎞
θc = sin −1 ⎛ ⎞ ≤ sin −1 ⎛
= 44.8°
⎝n ⎠
⎝ 1.42 ⎠
In the gasoline gauge, skylight from above travels down the plastic. The rays close to the vertical
are totally reflected from the sides of the slab and from both facets at the lower end of the plastic,
where it is not immersed in gasoline. This light returns up inside the plastic and makes it look
bright. Where the plastic is immersed in gasoline, with index of refraction about 1.50, total
internal reflection should not happen. The light passes out of the lower end of the plastic with
little reflected, making this part of the gauge look dark. To frustrate total internal reflection in the
gasoline, the index of refraction of the plastic should be n < 2.12 .
⎛ 1.50 ⎞
since
θ c = sin −1 ⎜
= 45.0°
⎝ 2.12 ⎟⎠
Additional Problems
*P35.40 From the textbook figure we have w = 2b + a
so
For refraction at entry,
w − a 700 µ m − 1 µ m
=
= 349.5 µ m
2
2
b 349.5 µ m
tan θ 2 = =
= 0.291
θ 2 = 16.2°
t 1 200 µ m
b=
n1 sin θ1 = n2 sin θ 2
n sin θ 2
1.55 sin 16.2°
= sin −1
= sin −1 0.433 = 25.7°
θ1 = sin −1 2
n1
1.00
310
P35.41
Chapter 35
Scattered light leaves the center of the photograph (a) in all horizontal
directions between θ1 = 0° and 90° from the normal. When it immediately
enters the water (b), it is gathered into a fan between 0° and θ 2 max
given by
q 1 max
(a)
n1 sin θ1 = n2 sin θ 2
1.00 sin 90 = 1.333 sin θ 2 max
q 2 max
θ 2 maax = 48.6°
(b)
The light leaves the cylinder without deviation, so the viewer only
receives light from the center of the photograph when he has turned by an
angle less than 48.6°. When the paperweight is turned farther, light at the
back surface undergoes total internal reflection (c). The viewer sees things
outside the globe on the far side.
(c)
FIG. P35.41
P35.42
Let n( x ) be the index of refraction at distance x below the top of the atmosphere and n( x = h) = n
be its value at the planet surface.
n( x ) = 1.000 + ⎛
⎝
Then,
(a)
The total time interval required to traverse the atmosphere is
h
∆t =
(b)
n − 1.000 ⎞
x
⎠
h
dx
∫0 v =
n ( x)
∫0 c dx :
∆t =
1 ⎡
⎛ n − 1.000 ⎞ ⎤
1.000 + ⎜
⎟⎠ x ⎥ dx
⎢
∫
⎝
c0⎣
h
⎦
∆t =
h ( n − 1.000 ) ⎛ h 2 ⎞
h ⎛ n + 1.000 ⎞
=
+
⎜
⎟⎠
⎟
⎜
2
c
ch
c⎝
⎝ 2⎠
h
h
h
The travel time in the absence of an atmosphere would be .
c
⎛ n + 1.000 ⎞
Thus, the time in the presence of an atmosphere is ⎜
⎟⎠ times larger .
⎝
2
P35.43
Let the air and glass be medium 1 and 2, respectively. By Snell’s law, n2 sin θ 2 = n1 sin θ1
or
1.56 sin θ 2 = sin θ1
But the conditions of the problem are such that θ1 = 2θ 2.
1.56 sin θ 2 = sin 2θ 2
We now use the double-angle trig identity suggested.
1.56 sin θ 2 = 2 sin θ 2 cos θ 2
or
cos θ 2 =
Thus, θ 2 = 38.7° and θ1 = 2θ 2 = 77.5° .
1.56
= 0.780
2
The Nature of Light and the Laws of Geometric Optics
P35.44
(a)
θ1′ = θ1 = 30.0°
311
n1 sin θ1 = n2 sin θ 2
1.00 sin 30.0° = 1.55 sin θ 2
θ 2 = 18.8°
(b)
θ1′ = θ1 = 30.0°
FIG. P35.44
⎛ n sin θ1 ⎞
θ 2 = sin ⎜ 1
⎝ n2 ⎟⎠
−1
⎛ 1.55 sin 30.0° ⎞
= sin −1 ⎜
⎟⎠ = 50.8°
⎝
1
(c), (d)
The other entries are computed similarly, and are shown in the table below.
(c) air into glass, angles in degrees
incidence
reflection
refraction
0
0
0
10.0
10.0
6.43
20.0
20.0
12.7
30.0
30.0
18.8
40.0
40.0
24.5
50.0
50.0
29.6
60.0
60.0
34.0
70.0
70.0
37.3
80.0
80.0
39.4
90.0
90.0
40.2
(d) glass into air, angles in degrees
incidence
reflection
refraction
0
0
0
10.0
10.0
15.6
20.0
20.0
32.0
30.0
30.0
50.8
40.0
40.0
85.1
50.0
50.0
none*
60.0
60.0
none*
70.0
70.0
none*
80.0
80.0
none*
90.0
90.0
none*
*total internal reflection
P35.45
1
3
=
4 3 4
For water,
sin θ c =
Thus
θ c = sin −1 ( 0.750 ) = 48.6°
and
d = 2 ⎡⎣(1.00 m ) tan θ c ⎤⎦
d = ( 2.00 m ) tan 48.6° = 2.27 m
P35.46
(a)
FIG. P35.45
We see the Sun moving from east to west across the sky. Its angular speed is
ω=
∆θ
2π rad
=
= 7.27 × 10 −5 rad s
∆t 86 400 s
The direction of sunlight crossing the cell from the window changes at this rate, moving
on the opposite wall at speed
v = rω = (2.37 m)(7.27 × 10 −5 rad s) = 1.72 × 10 −4 m s = 0.172 mm s
(b)
The mirror folds into the cell the motion that would occur in a room twice as wide:
v = rω = 2 ( 0.174 mm s ) = 0.345 mm s
(c), (d) As the Sun moves southward and upward at 50.0°, we may regard the corner of the
window as fixed, and both patches of light move northward and downward at 50.0° .
312
P35.47
Chapter 35
Horizontal light rays from the setting Sun
pass above the hiker. The light rays are
twice refracted and once reflected, as in
Figure (b). The most intense light reaching
the hiker, that which represents the visible
rainbow, is located between angles of 40°
and 42° from the hiker’s shadow. The hiker
sees a greater percentage of the violet inner
edge, so we consider the red outer edge.
The radius R of the circle of droplets is
Figure (a)
R = ( 8.00 km ) sin 42.0° = 5.35 km
Then the angle f, between the vertical and the radius
where the bow touches the ground, is given by
cos φ =
2.00 km 2.00 km
=
= 0.374
R
5.35 km
φ = 68.1°
or
The angle filled by the visible bow is 360° − ( 2 × 68.1° ) = 224°
224°
so the visible bow is
= 62.2% of a circle .
360°
P35.48
(a)
45.0°
(b)
Yes
Figure (b)
FIG. P35.47
as shown in the first figure to the right.
If grazing angle is halved, the number of reflections from
the side faces is doubled.
P35.49
FIG. P35.48
As the beam enters the slab,
1.00 sin 50.0° = 1.48 sin θ 2
giving
θ 2 = 31.2°
FIG. P35.49
1.55 mm
from the left end. Thereafter, the beam
tan 31.2°
strikes a face each time it has traveled a distance of 2 x1 along the length of the slab. Since the slab
is 420 mm long, the beam has an additional 420 mm − x1 to travel after the first reflection. The
number of additional reflections is
The beam then strikes the top of the slab at x1 =
420 mm − x1 420 mm − 1.55 mm tan 31.2°
=
= 81.5
2 x1
3.10 mm taan 31.2°
or 81 reflections
since the answer must be an integer. The total number of reflections made in the slab is then 82 .
The Nature of Light and the Laws of Geometric Optics
P35.50
Light passing the top of the pole makes an angle of incidence
φ1 = 90.0° − θ . It falls on the water surface at distance from the pole
L−d
s1 =
tan θ
and has an angle of refraction
Then
and the whole shadow length is
φ2 from 1.00 sin φ1 = n sin φ2
s2 = d tan φ2
s1 + s2 =
L−d
⎛
⎛ sin φ1 ⎞ ⎞
+ d tan ⎜ sin −1 ⎜
⎝ n ⎟⎠ ⎟⎠
⎝
tan θ
s1 + s2 =
L−d
⎛
⎛ cos θ ⎞ ⎞
+ d tan ⎜ sin −1 ⎜
⎝ n ⎟⎠ ⎟⎠
⎝
tan θ
=
P35.51
313
2.00 m
⎛
⎛ cos 40.0° ⎞ ⎞
+ ( 2.00 m ) tan ⎜ sin −1 ⎜
= 3.79 m
⎝ 1.33 ⎟⎠ ⎟⎠
⎝
tan 40.0°
FIG. P35.50
Define n1 to be the index of refraction of the surrounding medium and n2 to
be that for the prism material. We can use the critical angle of 42.0° to find
the ratio n2 :
n1
n2 sin 42.0° = n1 sin 90.0°
So,
n2
1
=
= 1.49
n1 sin 42.0°
Call the angle of refraction θ 2 at the surface 1. The ray inside the prism
FIG. P35.51
forms a triangle with surfaces 1 and 2, so the sum of the interior angles of
this triangle must be 180°.
Thus,
( 90.0° − θ 2 ) + 60.0° + ( 90.0° − 42.0°) = 180°
Therefore,
θ 2 = 18.0°
Applying Snell’s law at surface 1,
n1 sin θ1 = n2 sin 18.0°
⎛n ⎞
sin θ1 = ⎜ 2 ⎟ sin θ 2 = 1.49 sin 18.0°
⎝ n1 ⎠
*P35.52 (a)
θ1 = 27.5°
As the mirror turns through angle θ , the angle of incidence increases
by θ and so does the angle of reflection. The incident ray is stationary, so the reflected ray turns through angle 2θ . The angular speed of
the reflected ray is 2ω m . The speed of the dot of light on the circular
wall is 2ω m R = 2( 35 rad/s)(3 m) = 210 m/s .
(b)
The two angles marked θ in the figure to the right are equal because
their sides are perpendicular, right side to right side and left side to
left side.
d
ds
cos θ =
=
We have
2
2
dx
x +d
and
ds
= 2ω m x 2 + d 2 . So
dt
dx ds
=
dt dt
FIG. P35.52
x2 + d 2
x2 + d 2
x 2 + ( 3 m)2
x2 + 9 m2
= 2ω m
= 2( 35 rad/s)
= 23.3
d
d
m ⋅s
( 3 m)
continued on next page
314
Chapter 35
(c)
The minimum value is 210 m/s for x = 0, just as for the circular wall.
(d)
The maximum speed goes to infinity as x goes to infinity , which happens when the mirror
turns by 45°.
(e)
To turn by 45° takes the time interval (p /4 rad)/(35 rad/s) = 22.4 ms .
*P35.53 (a)
For polystyrene surrounded by air, internal reflection requires
⎛ 1.00 ⎞
= 42.2°
θ 3 = sin −1 ⎜
⎝ 1.49 ⎟⎠
Then from geometry,
θ 2 = 90.0° − θ 3 = 47.8°
From Snell’s law,
sin θ1 = 1.49 sin 47.8° = 1.10
This has no solution.
Therefore, total internal reflection always happens or the
greatest angle is 90°.
(b)
(c)
For polystyrene surrounded by water,
⎛ 1.33 ⎞
= 63.2°
θ 3 = sin ⎜
⎝ 1.49 ⎟⎠
and
θ 2 = 26.8°
From Snell’s law,
θ1 = 30.3°
FIG. P35.53
−1
No internal refraction is possible since the beam is initially traveling in a medium of lower
index of refraction. No angle exists.
*P35.54 (a)
The optical day is longer. Incoming sunlight is
q1
refracted downward at the top of the atmosphere, so
d
an observer can see the rising Sun when it is still geo1
metrically below the horizon. Light from the setting
2
Sun reaches her after the Sun is below the horizon
geometrically.
q2
f
FIG. P35.54
(b) The picture illustrates optical sunrise. At the center of
the Earth,
6.37 × 10 6 m
6.37 × 10 6 m + 8 614
φ = 2.98°
θ 2 = 90 − 2.98° = 87.0°
cos φ =
At the top of the atmosphere
n1 sin θ1 = n2 sin θ 2
1 sin θ1 = 1.000 293 sin 87.0°
θ1 = 87.4°
Deviation upon entry is
δ = θ1 − θ 2
δ = 87.364° − 87.022° = 0.342°
⎛ 86 400 s ⎞
Sunrise of the optical day is before geometric sunrise by 0.342° ⎜
= 82.2 s.
⎝ 360° ⎟⎠
Optical sunset occurs later too, so the optical day is longer by 164 s .
The Nature of Light and the Laws of Geometric Optics
P35.55
tan θ1 =
315
4.00 cm
h
and
tan θ 2 =
2.00 cm
h
tan 2 θ1 = ( 2.00 tan θ 2 ) = 4.00 tan 2 θ 2
2
⎛ sin 2 θ 2 ⎞
sin 2 θ1
=
4
.
00
⎜⎝ 1 − sin 2 θ ⎟⎠
1 − sin 2 θ1
2
Snell’s law in this case is:
(1)
n1 sin θ1 = n2 sin θ 2
sin θ1 = 1.333 sin θ 2
Squaring both sides,
sin 2 θ1 = 1.777 sin 2 θ 2
Substituting (2) into (1),
⎛ sin 2 θ 2 ⎞
1.777 sin 2 θ 2
4
.
00
=
⎜⎝ 1 − sin 2 θ ⎟⎠
1 − 1.777 sin 2 θ 2
2
Defining x = sin 2 θ ,
0.444
1
=
1 − 1.777 x 1 − x
Solving for x,
0.444 − 0.444 x = 1 − 1.777 x and x = 0.417
(2)
FIG. P35.55
From x we can solve for θ 2 : θ 2 = sin −1 0.417 = 40.2°
Thus, the height is
P35.56
δ = θ1 − θ 2 = 10.0°
h=
2.00 cm 2.00 cm
=
= 2.36 cm
tan θ 2
tan 40.2°
and
n1 sin θ1 = n2 sin θ 2
with
n1 = 1, n2 =
4
3
Thus,
θ1 = sin −1 ( n2 sin θ 2 ) = sin −1 ⎡⎣ n2 sin (θ1 − 10.0° ) ⎤⎦
(You can use a calculator to home in on an approximate solution to this equation, testing different
values of θ1 until you find that θ1 = 36.5° . Alternatively, you can solve for θ1 exactly, as shown
below.)
4
We are given that
sin θ1 = sin (θ1 − 10.0° )
3
3
This is the sine of a difference, so
sin θ1 = sin θ1 cos 10.0° − cos θ1 sin 10.0°
4
3⎞
⎛
Rearranging,
sin 10.0° cos θ1 = ⎜ cos 10.0° − ⎟ sin θ1
⎝
4⎠
sin 10.0°
= tan θ1 and
θ1 = tan −1 ( 0.740 ) = 36.5°
cos 10.0° − 0.750
316
P35.57
Chapter 35
Observe in the sketch that the angle of incidence at point P is g ,
and using triangle OPQ:
sin γ =
Also,
L
R
cos γ = 1 − sin 2 γ =
R 2 − L2
R
Applying Snell’s law at point P, 1.00 sin γ = n sin φ .
sin γ
L
=
n
nR
Thus,
sin φ =
and
cos φ = 1 − sin 2 φ =
FIG. P35.57
n 2 R 2 − L2
nR
From triangle OPS, φ + (α + 90.0° ) + ( 90.0° − γ ) = 180° or the angle of incidence at point S is
α = γ − φ . Then, applying Snell’s law at point S
gives
or
1.00 sin θ = n sin α = n sin (γ − φ )
⎡⎛ L ⎞
sin θ = n [ sin γ cos φ − cos γ sin φ ] = n ⎢⎜ ⎟
⎢⎣⎝ R ⎠
L
sin θ = 2 n 2 R 2 − L2 − R 2 − L2
R
(
and
*P35.58 (a)
)
⎡L
θ = sin −1 ⎢ 2
⎣R
(
(c)
P35.59
R 2 − L2 ⎛ L ⎞ ⎤
⎜⎝ ⎟⎠ ⎥
R
nR ⎥⎦
)
⎤
n 2 R 2 − L2 − R 2 − L2 ⎥
⎦
In the textbook figure we have r1 = a 2 + x 2 and r2 = b 2 + ( d − x ) . The speeds in the two
media are v1 ⫽ c n1 and v2 ⫽ c n2 so the travel time for the light from P to Q is indeed
2
2
r r
n a 2 + x 2 n2 b + ( d − x )
t= 1 + 2 = 1
+
v1 v2
c
c
(b)
n 2 R 2 − L2
−
nR
2
−1 2
−1 2
dt
n
n
= 1 ( a 2 + x 2 ) 2 x + 2 ( b 2 + (d − x )2 ) 2(d − x )(−1) = 0 is the requirement
dx 2c
2c
n2 ( d − x )
n1 x
=
for minimal travel time, which simplifies to
.
2
2
2
a +x
b2 + (d − x )
Now
Now sin θ1 =
x
a +x
2
2
and sinθ 2 =
d−x
b + (d − x)
2
2
so we have directly n1 sin θ1 = n2 sin θ 2.
To derive the law of reflection, locate point O so that the time of
travel from point A to point B will be minimum.
The total light path is L = a sec θ1 + b sec θ 2 .
⎛1⎞
The time of travel is t = ⎜ ⎟ ( a sec θ1 + b sec θ 2 ) .
⎝v⎠
If point O is displaced by dx, then
⎛1⎞
dt = ⎜ ⎟ ( a sec θ1 tan θ1dθ1 + b sec θ 2 tan θ 2 dθ 2 ) = 0
⎝v⎠
(since for minimum time dt = 0).
continued on next page
FIG. P35.59
(1)
The Nature of Light and the Laws of Geometric Optics
Also,
c + d = a tan θ1 + b tan θ 2 = constant
so,
a sec 2 θ1dθ1 + b sec 2 θ 2 dθ 2 = 0
317
(2)
Divide equations (1) and (2) to find θ1 = θ 2 .
P35.60
As shown in the sketch, the angle of incidence at point A is:
⎛ d 2⎞
⎛ 1.00 m ⎞
θ = sin −1 ⎜
= sin −1 ⎜
= 30.0°
⎝ R ⎟⎠
⎝ 2.00 m ⎟⎠
If the emerging ray is to be parallel to the incident ray, the
path must be symmetric about the centerline CB of the
cylinder. In the isosceles triangle ABC,
γ =α
β = 180° − θ
and
Therefore,
α + β + γ = 180°
FIG. P35.60
2α + 180° − θ = 180°
θ
α = = 15.0°
or
2
Then, applying Snell’s law at point A,
becomes
n sin α = 1.00 sin θ
n=
or
P35.61
(a)
sin θ sin 30.0°
=
= 1.93
sin α sin 15.0°
The apparent radius of the glowing sphere is R3 as
shown. For it
R
sin θ1 = 1
R2
R
sin θ 2 = 3
R2
n sin θ1 = 1 sin θ 2
n
R1 R3
=
R2 R2
R3 = nR1
q2
q1
R3
R1
R2
FIG. P35.61 (a)
(b)
nR1
.
R2
The ray considered in part (a) undergoes total internal
reflection. In this case a ray escaping the
atmosphere as shown here is responsible for the
apparent radius of the glowing sphere and
If nR1 > R2, then sin θ 2 cannot be equal to
R3 = R2 .
R3
FIG. P35.61 (b)
318
P35.62
Chapter 35
(a)
At the boundary of the air and glass, the critical angle is given by
1
sin θ c =
n
d 4
sin θ c
d
= .
Consider the critical ray PBB′: tan θ c =
or
B'
t
cos θ c 4 t
B
air
n
t
qc
P
d/4
2
Squaring the last equation gives:
Since sin θ c =
(b)
(c)
1
, this becomes
n
sin 2 θ c
sin 2 θ c
⎛ d⎞
=
=⎜ ⎟ .
cos 2 θ c 1 − sin 2 θ c ⎝ 4 t ⎠
2
d
FIG. P35.62
1
⎛ d⎞
4t
= ⎜ ⎟ or n = 1 + ⎛⎜ ⎞⎟
⎝ d⎠
n − 1 ⎝ 4t ⎠
2
2
Solving for d,
d=
Thus, if n = 1.52 and t = 0.600 cm,
d=
4t
n2 − 1
.
4 ( 0.600 cm )
(1.52 )2 − 1
= 2.10 cm
Since violet light has a larger index of refraction, it will lead to a smaller critical angle and
the inner edge of the white halo will be tinged with violet light.
P35.63
(a)
Given that
θ1 = 45.0° and θ 2 = 76.0°
Snell’s law at the first surface gives
n sin α = 1.00 sin 45.0°
(1)
Observe that the angle of incidence at the second
surface is
β = 90.0° − α .
Thus, Snell’s law at the second surface yields
n sin β = n sin ( 90.0° − α ) = 1.00 sin 76.0°
n cos α = sin 76.0°.
or
(b)
FIG. P35.63
(2)
sin 45.0°
= 0.729
sin 76.0°
Dividing Equation (1) by Equation (2),
tan α =
or
α = 36.1°
Then, from Equation (1),
n=
sin 45.0° sin 45.0°
=
= 1.20
sin α
sin 36.1°
L
From the sketch, observe that the distance the light travels in the plastic is d =
. Also,
sin α
c
the speed of light in the plastic is v = , so the time required to travel through the plastic is
n
∆t =
1.20 ( 0.500 m )
d
nL
= 3.40 × 10 −9 s = 3.40 ns
=
=
v c sin α ( 3.00 × 108 m s ) sin 36.1°
The Nature of Light and the Laws of Geometric Optics
*P35.64
sin θ1
sin θ 2
sin θ1
sin θ 2
0.174
0.342
0.500
0.643
0.7666
0.866
0.940
0.985
0.131
0.261
0.379
0.480
0.576
0.647
0.711
0.740
1.330 4
1.312 9
1.317 7
1.338 5
1.3228 9
1.339 0
1.322 0
1.331 5
The straightness of the graph line demonstrates Snell’s
proportionality of the sine of the angle of refraction to the
sine of the angle of incidence.
The slope of the line is
n = 1.327 6 ± 0.01
319
FIG. P35.64
The equation sin θ1 = n sin θ 2 shows that this slope is the index of refraction,
n = 1.328 ± 0.8%
*P35.65 Consider an insulated box with the imagined one-way mirror forming one face, installed so that
90% of the electromagnetic radiation incident from the outside is transmitted to the inside and
only a lower percentage of the electromagnetic waves from the inside make it through to the outside. Suppose the interior and exterior of the box are originally at the same temperature. Objects
within and without are radiating and absorbing electromagnetic waves. They would all maintain
constant temperature if the box had an open window. With the glass letting more energy in than
out, the interior of the box will rise in temperature. But this is impossible, according to Clausius’s
statement of the second law. This reduction to a contradiction proves that it is impossible for the
one-way mirror to exist.
ANSWERS TO EVEN PROBLEMS
P35.2
227 Mm s
P35.4
(a) and (b) See the solution.
P35.6
See the solution.
P35.8
27.1 ns
P35.10
15.4°, 2.56 m. The light wave slows down as it moves from air to water but the sound wave
speeds up by a large factor. The light wave bends toward the normal and its wavelength shortens,
but the sound wave bends away from the normal and its wavelength increases.
P35.12
(a) 474 THz
P35.14
(a) 2.0 × 108 m/s
P35.16
6.30 cm
P35.18
3.39 m
(b) 422 nm
(c) 0.055 7°
(c) 200 Mm s
(b) 474 THz
(c) 4.2 × 10−7 m
320
Chapter 35
P35.20
106 ps
P35.22
23.1°
P35.24
∼10−11 s; between 103 and 104 wavelengths
P35.26
(a) 41.5° (b) 18.5° (c) 27.6° (d) 42.6°
P35.28
sin −1
P35.30
See the solution.
P35.32
0.171°
P35.34
62.5°
P35.36
67.2°
P35.38
(a) 10.7° (b) air (c) Sound falling on the wall from most directions is 100% reflected.
P35.40
25.7°
P35.42
(a)
P35.44
(a) See the solution; the angles are θ1′ = 30.0° and θ 2 = 18.8° (b) See the solution; the angles are
θ1′ = 30.0° and θ 2 = 50.8°. (c) and (d) See the solution.
P35.46
(a) 0.172 mm/s (b) 0.345 mm/s (c) Northward at 50.0° below the horizontal
(d) Northward at 50.0° below the horizontal
P35.48
(a) 45.0°
P35.50
3.79 m
P35.52
(a) 210 m/s (b) 23.3 (x2 + 9 m2)/m⋅s (c) 210 m/s for x = 0; this is the same as the speed on the
circular wall. (d) The speed goes to infinity as x goes to infinity, when the mirror turns through
45° from the x = 0 point. (e) 22.4 ms
P35.54
(a) The optical day is longer. Incoming sunlight is refracted downward at the top of the atmosphere,
so an observer can see the rising Sun when it is still geometrically below the horizon. Light from
the setting Sun reaches her after the Sun is already below the horizon geometrically.
(b) 164 s
P35.56
36.5°
P35.58
See the solution.
P35.60
1.93
P35.62
(a) n = [1 + (4t/d)2]1/2
P35.64
See the solution; n = 1.328 ± 0.8%.
(
n 2 − 1 sin Φ − cos Φ
h ⎛ n + 1⎞
⎜
⎟
c⎝ 2 ⎠
)
(b) larger by
n +1
times
2
(b) Yes; see the solution.
(b) 2.10 cm
(c) violet
36
Image Formation
CHAPTER OUTLINE
36.1
36.2
Images Formed by Flat Mirrors
Images Formed by Spherical
Mirrors
36.3 Images Formed by Refraction
36.4 Thin Lenses
36.5 Lens Aberrations
36.6 The Camera
36.7 The Eye
36.8 The Simple Magnifier
36.9 The Compound Microscope
36.10 The Telescope
ANSWERS TO QUESTIONS
Q36.1
With a concave spherical mirror, for objects beyond the
focal length the image will be real and inverted. For
objects inside the focal length, the image will be virtual,
upright, and magnified. Try a shaving or makeup mirror
as an example.
Q36.2
With a convex spherical mirror, all images of real objects
are upright, virtual and smaller than the object. As seen
in Question 36.1, you only get a change of orientation
when you pass the focal point—but the focal point of a
convex mirror is on the non-reflecting side!
*Q36.3 (i) When we flatten a curved mirror we move its center of curvature out to infinity. The focal
length is still half the radius of curvature and is infinite. Answer (d).
(ii) The image is actual size and right side up. The magnification is 1. Answer (b).
Q36.4
The mirror equation and the magnification equation apply to plane mirrors. A curved mirror is
made flat by increasing its radius of curvature without bound, so that its focal length goes to
1
1
1 1 1
infinity. From + = = 0 we have = − ; therefore, p = − q. The virtual image is as far
p
q
p q f
q p
behind the mirror as the object is in front. The magnification is M = − = = 1. The image is
p p
right side up and actual size.
*Q36.5 (i) Answer (c). (ii) Answer (c). When the object is at the focal point the image can be thought of as
a right side up image behind the mirror at infinity, or as an inverted image in front of the mirror.
Q36.6
In the diagram, only two of the three principal rays have
been used to locate images to reduce the amount of visual
clutter. The upright shaded arrows are the objects, and the
correspondingly numbered inverted arrows are the images.
As you can see, object 2 is closer to the focal point than
object 1, and image 2 is farther to the left than image 1.
O1
O2
F
C
V
I2 I1
FIG. Q36.6
321
322
Chapter 35
36
*Q36.7 (i) (a) positive (b) negative (c) negative (d) negative (e) positive (f) positive (g) positive
(ii) (a) positive (b) positive (c) positive (d) positive (e) negative (f) negative
(iii) (a) positive (b) negative (c) positive (d) negative (e) negative (f) positive
*Q36.8 Answer (c). The angle of refraction for the light coming from fish to person is 60°. The angle of
incidence is smaller, so the fish is deeper than it appears.
*Q36.9 The ranking is e > d > g > a > b > f > c. In case e, the object is at infinite distance. In d the object
distance is very large but not infinite. In g the object distance is several times the focal length. In
a, the object distance is a little larger than the focal length. In b the object distance is very slightly
larger than the focal length. In f it is equal to the focal length. In c the object distance is less than
the focal length.
Q36.10 An infinite number. In general, an infinite number of
rays leave each point of any object and travel in all
directions. Note that the three principal rays that we
use for imaging are just a subset of the infinite number
of rays. All three principal rays can be drawn in a ray
diagram, provided that we extend the plane of the lens as
shown in Figure Q36.10.
F
O
I
F
FIG. Q36.10
*Q36.11 Answer (d). The entire image is visible, but only at half the intensity. Each point on the object is
a source of rays that travel in all directions. Thus, light from all parts of the object goes through
all unblocked parts of the lens and forms an image. If you block part of the lens, you are blocking
some of the rays, but the remaining ones still come from all parts of the object.
*Q36.12 Answer (e). If the object distance is 2f the image distance is also 2f and the distance between
object and real image is minimal.
*Q36.13 The focal point is defined as the location of the image formed by rays originally parallel to the
axis. An object at a large but finite distance will radiate rays nearly but not exactly parallel.
Infinite object distance describes the definite limiting case in which these rays become parallel.
To measure the focal length of a converging lens, set it up to form an image of the farthest object
you can see outside a window. The image distance will be equal to the focal length within one
percent or better if the object distance is a hundred times larger or more.
*Q36.14 Use a converging lens as the projection lens in a slide projector. Place the brightly illuminated
slide slightly farther than its focal length away from it, so that the lens will produce a real,
inverted, enlarged image on the screen.
*Q36.15 Answer (e). The water drop functions as a lens of short focal length, forming a real image of the
distant object in space, outside the drop on the side where the light exits the drop. The camera
lens is focused on the real image.
Q36.16 Chromatic aberration arises because a material medium’s refractive index can be frequency
dependent. A mirror changes the direction of light by reflection, not refraction. Light of all
wavelengths follows the same path according to the law of reflection, so no chromatic aberration
happens.
Image Formation
323
Q36.17 If the converging lens is immersed in a liquid with an index of refraction significantly greater
than that of the lens itself, it will make light from a distant source diverge. This is not the case
with a converging (concave) mirror, as the law of reflection has nothing to do with the indices of
refraction.
Q36.18 As in the diagram, let the center of curvature C of the
fishbowl and the bottom of the fish define the optical axis,
intersecting the fishbowl at vertex V. A ray from the top
of the fish that reaches the bowl surface along a radial
line through C has angle of incidence zero and angle of
refraction zero. This ray exits from the bowl unchanged in
direction. A ray from the top of the fish to V is refracted
to bend away from the normal. Its extension back inside
the fishbowl determines the location of the image and the
characteristics of the image. The image is upright, virtual,
and enlarged.
V
C
O I
FIG. Q36.18
Q36.19 Because when you look at the ECNALUBMA in your rear view mirror, the apparent left-right
inversion clearly displays the name of the AMBULANCE behind you. Do not jam on your brakes
when a MIAMI city bus is right behind you.
Q36.20 With the meniscus design, when you direct your gaze near the outer circumference of the lens
you receive a ray that has passed through glass with more nearly parallel surfaces of entry and
exit. Thus, the lens minimally distorts the direction to the object you are looking at. If you wear
glasses, turn them around and look through them the wrong way to maximize this distortion.
Q36.21 Answer (b). The outer surface should be flat so that it will not produce a fuzzy or distorted image
for the diver when the mask is used either in air or in water.
Q36.22 The eyeglasses on the left are diverging lenses that correct for nearsightedness. If you look
carefully at the edge of the person’s face through the lens, you will see that everything viewed
through these glasses is reduced in size. The eyeglasses on the right are converging lenses, which
correct for farsightedness. These lenses make everything that is viewed through them look larger.
Q36.23 The eyeglass wearer’s eye is at an object distance from the lens that is quite small—the eye is on
the order of 10 −2 meter from the lens. The focal length of an eyeglass lens is several decimeters,
positive or negative. Therefore the image distance will be similar in magnitude to the object
distance. The onlooker sees a sharp image of the eye behind the lens. Look closely at the left
side of Figure Q36.22 and notice that the wearer’s eyes seem not only to be smaller, but also
positioned a bit behind the plane of his face—namely where they would be if he were not
wearing glasses. Similarly, in the right half of Figure Q36.22, his eyes seem to be in front of the
plane of his face and magnified. We as observers take this light information coming from the
object through the lens and perceive or photograph the image as if it were an object.
Q36.24 Absolutely. Only absorbed light, not transmitted light, contributes internal energy to a transparent
object. A clear lens can stay ice-cold and solid as megajoules of light energy pass through it.
Q36.25 Make the mirror an efficient reflector (shiny). Make it reflect to the image even rays far from the
axis, by giving it a parabolic shape. Most important, make it large in diameter to intercept a lot of
solar power. And you get higher temperature if the image is smaller, as you get with shorter focal
length; and if the furnace enclosure is an efficient absorber (black).
324
Chapter 36
Q36.26 The artist’s statements are accurate, perceptive, and eloquent. The image you see is “almost one’s
whole surroundings,” including things behind you and things farther in front of you than the
globe is, but nothing eclipsed by the opaque globe or by your head. For example, we cannot see
Escher’s index and middle fingers or their reflections in the globe.
The point halfway between your eyes is indeed the focus in a figurative sense, but it is not an
optical focus. The principal axis will always lie in a line that runs through the center of the sphere
and the bridge of your nose. Outside the globe, you are at the center of your observable universe.
If you wink at the ball, the center of the looking-glass world hops over to the location of the
image of your open eye.
Q36.27 You have likely seen a Fresnel mirror for
sound. The diagram represents first a side
view of a band shell. It is a concave mirror for
sound, designed to channel sound into a beam
toward the audience in front of the band shell.
Sections of its surface can be kept at the right
orientations as they are pushed around inside
a rectangular box to form an auditorium with
good diffusion of sound from stage to audience,
with a floor plan suggested by the second part
of the diagram.
FIG. Q36.27
SOLUTIONS TO PROBLEMS
Section 36.1
P36.1
Images Formed by Flat Mirrors
I stand 40 cm from my bathroom mirror. I scatter light, which travels to the mirror and back to
me in time
0.8 m
~10 −9 s
3 × 108 m s
showing me a view of myself as I was at that look-back time. I’m no Dorian Gray!
P36.2
The virtual image is as far behind the mirror as the choir is in
front of the mirror. Thus, the image is 5.30 m behind the mirror. The image of the choir is 0.800 m + 5.30 m = 6.10 m from
the organist. Using similar triangles:
h′
6.10 m
=
0.600 m 0.800 m
or
6.10 m ⎞
h ′ = ( 0.600 m ) ⎛
= 4.58 m
⎝ 0.800 m ⎠
View Looking Down
South
image of choir
mirror
0.600 m
Organist
0.800 m
5.30 m
FIG. P36.2
h'
Image Formation
P36.3
325
The flatness of the mirror is described
by
R = ∞, f = ∞
and
1
=0
f
By our general mirror equation,
FIG. P36.3
1 1 1
+ = =0
p q f
or
q = −p
Thus, the image is as far behind the mirror as the person is in front. The magnification is then
M=
so
h′
−q
=1=
p
h
h ′ = h = 70.0 inches
The required height of the mirror is defined by the triangle from the person’s eyes to the top and
bottom of his image, as shown. From the geometry of the triangle, we see that the mirror height
must be:
⎛ p ⎞ h′
⎛ p ⎞
h′ ⎜
= h′ ⎜ ⎟ =
⎟
⎝ 2p⎠ 2
⎝ p − q⎠
Thus, the mirror must be at least 35.0 inches high .
P36.4
(1)
The first image in the left mirror is 5.00 ft behind the mirror, or 10.0 ft from the position
of the person.
(2)
The first image in the right mirror is located 10.0 ft behind the right mirror, but this location
is 25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind the
mirror, or 30.0 ft from the person.
(3)
The first image in the left mirror forms an image in the right mirror. This first image is
20.0 ft from the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed.
This image in the right mirror also forms an image in the left mirror. The distance from this
image in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is,
thus, 35.0 ft behind the mirror, or 40.0 ft from the person.
326
P36.5
Chapter 36
(a)
The flat mirrors have
R→∞
and
f →∞
The upper mirror M1
produces a virtual, actual sized
image I1 according to
1 1 1 1
+ = = =0
p1 q1 f ∞
q1 = − p1
with
M1 = −
q1
= +1
p1
As shown, this image is above the
upper mirror. It is the object for
mirror M 2 , at object distance
FIG. P36.5
p2 = p1 + h
The lower mirror produces
a virtual, actual-size, rightside-up image according to
1
1
+
=0
p2 q2
q2 = − p2 = − ( p1 + h )
with
M2 = −
q2
= +1 and M overall = M 1 M 2 = 1.
p2
Thus the final image is at distance p1 + h behind the lower mirror.
(b)
(c)
It is virtual .
Upright
(d)
With magnification +1 .
(e)
It does not appear to be reversed left and right. In a top view of the periscope, parallel
rays from the right and left sides of the object stay parallel and on the right and left.
Image Formation
Section 36.2
P36.6
327
Images Formed by Spherical Mirrors
For a concave mirror, both R and f are positive.
We also know that
(a)
f =
R
= 10.0 cm
2
1 1 1
1
1
3
= − =
−
=
q f p 10.0 cm 40.0 cm 40.0 cm
and
q = 13.3 cm
M=
q
13.3 cm
=−
= −0.333
p
40.0 cm
The image is 13.3 cm in front of the mirror, real, and inverted .
(b)
1 1 1
1
1
1
= − =
−
=
q f p 10.0 cm 20.0 cm 20.0 cm
and
q = 20.0 cm
M=
q
20.0 cm
=−
= −1.00
p
20.0 cm
The image is 20.0 cm in front of the mirror, real, and inverted .
(c)
1 1 1
1
1
= − =
−
=0
q f p 10.0 cm 10.0 cm
Thus,
q = infinity
No image is formed . The rays are reflected parallel to each other.
P36.7
(a)
1 1 2
+ =
p q R
gives
1
2
1
=−
−
= −0.083 3 cm −1 so
q
40.0 cm 30.0 cm
1
1
2
+ =
30.0 cm q ( −40.0 cm )
q = −12.0 cm
M=
(b)
1 1 2
+ =
p q R
gives
1
2
1
=−
−
= −0.066 6 cm −1 so
q
40.0 cm 60.0 cm
1
1
2
+ =
60.0 cm q ( −40.0 cm )
q = −15.0 cm
M=
(c)
Since M > 0, the images are upright .
−q
( −12.0 cm )
=−
= 0.400
p
30.0 cm
−q
( −15.0 cm )
=−
= 0.250
p
60.0 cm
328
P36.8
Chapter 36
1 1 1
1
1
= − =−
−
q f p
0.275 m 10.0 m
q = −0.267 m
gives
Thus, the image is virtual .
M=
−q
−0.267
=−
= 0.026 7
p
10.0 m
Thus, the image is upright ( + M )
P36.9
(a)
and
diminished
(M
1 1 2
+ =
p q R
becomes
1
2
1
=
−
q 60.0 cm 90.0 cm
q = 45.0 cm
and
M=
becomes
1
2
1
=
−
q 60.0 cm 20.0 cm
−q
( −60.0 cm )
M=
=−
= 3.00
p
( 20.0 cm )
(b) 1 + 1 = 2
p q R
q = −60.0 cm
and
−q
45.0 cm
=−
= −0.500
p
90.0 cm
(c) The image (a) is real, inverted, and diminished. That of (b) is
virtual, upright, and enlarged. The ray diagrams are similar to
Figures 36.13(a) and 36.13(b) in the text, respectively.
P36.10
< 1)
FIG. P36.9
With radius 2.50 m, the cylindrical wall is a highly efficient mirror for sound, with focal length
f =
R
= 1.25 m
2
In a vertical plane the sound disperses as usual, but that radiated in a horizontal plane is concentrated in a sound image at distance q from the back of the niche, where
1 1 1
+ =
p q f
so
1
1
1
+ =
2.00 m q 1.25 m
q = 3.33 m
*P36.11 (a)
Since the object is in front of the mirror, p > 0. With the image behind the mirror, q < 0. The
mirror equation gives the radius of curvature as
or R = 2 ⎛
⎝
(b)
2 1 1
1
1
10 − 1
= + =
−
=
,
R p q 1.00 cm 10.0 cm 10.0 cm
10.0 cm ⎞
= +2.22 cm .
9 ⎠
q
( −10.0 cm )
= +10.0 .
The magnification is M = − = −
p
1.00 cm
Image Formation
P36.12
329
The ball is a convex mirror with R = −4.25 cm
R
= −2.125 cm. We have
2
3
q
M = =−
4
p
3
q=− p
4
O
1 1 1
+ =
p q f
FIG. P36.12
and f =
I F
1
1
1
+
=
p − ( 3 4 ) p −2.125 cm
3
4
1
−
=
3p 3p −2.125 cm
3p = 2.125 cm
p = 0.708 cm in front of the sphere.
The image is upright, virtual, and diminished.
P36.13
(a)
(b)
M = −4 = −
q
p
q = 4p
q − p = 0.60 m = 4 p − p
p = 0.2 m
1 1 1
1
1
= + =
+
f p q 0.2 m 0.8 m
f = 160 mm
M =+
q
1
=−
p
2
q = 0.8 m
p = −2q
q + p = 0.20 m = − q + p = − q − 2q
P36.14
q = −66.7 mm
p = 133 mm
1 1 2
1
1
+ = =
+
p q R 0.133 m − 0.066 7 m
R = −267 mm
M =−
q
p
q = − Mp = −0.013 ( 30 cm ) = −0.39 cm
1 1 1 2
+ = =
p q f R
1
1
2
+
=
30 cm −0.39 cm R
2
R=
= −0.790 cm
−2.53 m −1
The cornea is convex, with radius of curvature 0.790 cm .
FIG. P36.14
C
330
P36.15
Chapter 36
With
M=
h ′ +4.00 cm
q
=
= +0.400 = −
h
10.0 cm
p
q = −0.400 p
the image must be virtual.
(a)
It is a convex mirror that produces a diminished upright virtual image.
(b)
We must have
p + q = 42.0 cm = p − q
p = 42.0 cm + q
p = 42.0 cm − 0.400 p
p=
42.0 cm
= 30.0 cm
1.40
The mirror is at the 30.0 cm mark .
(c)
1 1 1
1
1
1
= = − 0.050 0 cm
+ = =
+
p q f 30 cm −0.4 ( 30 cm ) f
f = −20.0 cm
The ray diagram looks like Figure 36.13(c) in the text.
P36.16
Assume that the object distance is the same in both cases (i.e., her face is the same distance from
the hubcap regardless of which way it is turned). Also realize that the near image (q = −10.0 cm)
occurs when using the convex side of the hubcap. Applying the mirror equation to both cases gives:
(concave side: R = R ,
q = −30.0 cm)
1
1
2
−
=
p 30.0
R
2
30.0 cm − p
=
R
( 30.0 cm ) p
or
(1)
(convex side: R = − R , q = −10.0 cm)
1
1
2
−
=−
p 10.0
R
2
p − 10.0 cm
=
R
(10.0 cm ) p
or
(a)
Equating Equations (1) and (2) gives:
30.0 cm − p
= p − 10.0 cm
3.00
or
p = 15.0 cm
Thus, her face is 15.0 cm from the hubcap.
continued on next page
(2)
Image Formation
(b)
Using the above result ( p = 15.0 cm ) in Equation (1) gives:
2
30.0 cm − 15.0 cm
=
R
( 30.0 cm ) (15.0 cm )
or
2
1
=
R
30.0 cm
and
R = 60.0 cm
The radius of the hubcap is 60.0 cm .
P36.17
(a)
q = ( p + 5.00 m ) and, since the image must be real,
M =−
q
= −5
p
q = 5p
or
p + 5.00 m = 5 p
Therefore,
or
p = 1.25 m
From
1 1 2
+ = ,
p q R
and
q = 6.25 m
R=
2 pq 2 (1.25) ( 6.25)
=
1.25 + 6.25
p+q
FIG. P36.17
= 2.08 m ( concavve )
P36.18
(b)
From part (a), p = 1.25 m; the mirror should be 1.25 m in front of the object.
(a)
The flat mirror produces an image according to
1 1 1 2
+ = =
p q f R
1
1 1
+ = =0
24 cm q ∞
q = −24.0 m
The image is 24.0 m behind the mirror, distant from your eyes by
1.55 m + 24.0 m = 25.6 m
(b)
The image is the same size as the object, so
θ=
h 1.50 m
=
= 0.058 7 rad
d 25.6 m
(c)
1 1 2
+ =
p q R
q=
1
= −0.960 m
− (1 1 m ) − (1 24 m )
1
1
2
+ =
24 m q ( −2 m )
This image is distant from your eyes by
(d)
The image size is given by M =
So its angular size at your eye is
(e)
h′
q
=−
h
p
1.55 m + 0.960 m = 2.51 m
q
−0.960 m ⎞
= −1.50 m ⎛
⎝ 24 m ⎠
p
= 0.060 0 m.
h ′ 0.06 m
θ′ = =
= 0.023 9 rad
d 2.51 m
h′ = −h
Your brain assumes that the car is 1.50 m high and calculates its distance as
d′ =
h 1.50 m
=
= 62.8 m
θ ′ 0.023 9
331
332
P36.19
Chapter 36
(a)
The image starts from a point whose height above the mirror vertex is given by
1 1 1 2
+ = =
p q f R
1
1
1
+ =
3.00 m q 0.500 m
Therefore,
q = 0.600 m
As the ball falls, p decreases and q increases. Ball and image pass when q1 = p1. When this
is true,
1
1
1
2
+
=
=
p1 p1 0.500 m p1
or
p1 = 1.00 m
As the ball passes the focal point, the image switches from infinitely far above the mirror
to infinitely far below the mirror. As the ball approaches the mirror from above, the virtual
image approaches the mirror from below, reaching it together when p2 = q2 = 0.
(b)
The falling ball passes its real image when it has fallen
3.00 m − 1.00 m = 2.00 m =
1 2
gt , or when t =
2
2 ( 2.00 m )
= 0.639 s
9.80 m s 2
The ball reaches its virtual image when it has traversed
3.00 m − 0 = 3.00 m =
Section 36.3
P36.20
1 2
gt , or at t =
2
2 ( 3.00 m )
= 0.782 s
9.80 m s 2
Images Formed by Refraction
When R → ∞ , the equation describing image formation at a single refracting surface becomes
⎛n ⎞
q = − p ⎜ 2 ⎟ . We use this to locate the final images of the two surfaces of the glass plate. First,
⎝ n1 ⎠
find the image the glass forms of the bottom of the plate.
1.33 ⎞
qB1 = − ⎛
(8.00 cm ) = −6.41 cm
⎝ 1.66 ⎠
This virtual image is 6.41 cm below the top surface of the glass of 18.41 cm below the water
surface. Next, use this image as an object and locate the image the water forms of the bottom of
the plate.
1.00 ⎞
qB2 = − ⎛
(18.41 cm ) = −13.84 cm
⎝ 1.33 ⎠
or
13.84 cm below the water surface
Now find image the water forms of the top surface of the glass.
1 ⎞
q3 = − ⎛
(12.0 cm ) = −9.02 cm
⎝ 1.33 ⎠
or
9.02 cm below the water surface
Therefore, the apparent thickness of the glass is ∆t = 13.84 cm − 9.02 cm = 4.82 cm .
P36.21
n1 n2 n2 − n1
+
=
= 0 and R → ∞
p q
R
1
n
q=− 2 p=−
( 50.0 cm ) = −38.2 cm
1.309
n1
Thus, the virtual image of the dust speck is 38.2 cm below the top surface of the ice.
Image Formation
P36.22
P36.23
n1 n2 n2 − n1
+
=
p q
R
333
1.00 1.50 1.50 − 1.00
1
+
=
=
p
q
6.00 cm
12.0 cm
becomes
(a)
1.00
1.50
1
+
=
20.0 cm
q
12.0 cm
or
q=
1.50
= 45.0 cm
⎡⎣(1.00 12.0 cm ) − (1.00 20.0 cm ) ⎤⎦
(b)
1.00
1.50
1
+
=
10.0 cm
q
12.0 cm
or
q=
1.50
= −90.0 cm
⎡⎣(1.00 12.0 cm ) − (1.00 10.0 cm ) ⎤⎦
(c)
1.00
1.50
1
+
=
3.0 cm
q
12.0 cm
or
q=
1.50
= −6.00 cm
⎡⎣(1.00 12.0 cm ) − (1.00 3.0 cm ) ⎤⎦
From Equation 36.8
Solve for q to find
n1 n2 n2 − n1
+
=
p q
R
n2 Rp
q=
p ( n2 − n1 ) − n1 R
In this case,
n1 = 1.50 , n2 = 1.00, R = −15.0 cm
and
p = 10.0 cm
So
q=
Therefore, the
(1.00 ) ( −15.0 cm ) (10.0 cm )
= −8.57 cm
(10.0 cm ) (1.00 − 1.50 ) − (1.50 ) ( −15.0 cm )
apparent depth is 8.57 cm
*P36.24 In the right triangle lying between O and the center of the curved surface, tan q1 = h/p. In the
right triangle lying between I and the center of the surface, tan q2 = −h′/q. We need the negative
sign because the image height is counted as negative while the angle is not. We substitute into the
given n1 tan q1 = n2 tan q2 to obtain n1 h/p = −n2 h′/q. Then the magnification, defined by
M = h′/h, is given by M = h′/h = −n1 q/n2 p.
*P36.25 (a)
The center of curvature is on the object side, so the radius of curvature is negative.
1.33 1.00 1.00 − 1.33
n1 n2 n2 − n1
+
=
q = −24.9 cm
becomes
+
=
30 cm
q
−80 cm
p q
R
So the image is inside the tank, 24.9 cm behind the front wall; virtual, right side up, enlarged.
(b)
Now we have
1.33 1.00 1.00 − 1.33
+
=
90 cm
q
−80 cm
q = −93.9 cm
So the image is inside the tank, 93.9 cm behind the front wall; virtual, right side up, enlarged.
(c)
In case (a) the result of problem 24 gives M = −
In case (b) we have M = −
(d)
n1q
1.33 ( −24.9 )
=−
= +1.10
n2 p
1.00(30)
1.33 ( −93.9 )
= +1.39
1.00(90)
In case (a) h′ = Mh = 1.10(9.00 cm) = 9.92 cm . In case (b), the farther lobster looms larger:
h′ = Mh = 1.30(9.00 cm) = 12.5 cm
(e)
The plastic has uniform thickness, so the surfaces of entry and exit for any particular ray
are very nearly parallel. The ray is slightly displaced, but it would not be changed in direction by going through the plastic wall with air on both sides. Only the difference between
the air and water is responsible for the refraction of the light.
334
P36.26
Chapter 36
For a plane surface,
n1 n2 n2 − n1
n p
+
=
becomes q = − 2
p q
R
n1
Thus, the magnitudes of the rate of change in the image and object positions are related by
dq n2 dp
=
dt
n1 dt
If the fish swims toward the wall with a speed of 2.00 cm s, the speed of the image is given by
dq 1.00
vimage =
=
( 2.00 cm s ) = 1.50 cm s
dt 1.33
Section 36.4
P36.27
(a)
Thin Lenses
⎡1
⎡ 1
⎤
1
1 ⎤
1
−
= ( n − 1) ⎢ − ⎥ = ( 0.440 ) ⎢
⎥
f
⎣ 12.0 cm ( −18.0 cm ) ⎦
⎣ R1 R2 ⎦
f = 16.4 cm
(b)
⎡ 1
⎤
1
1
= ( 0.440 ) ⎢
−
⎥
)
f
18
0
12
0
.
cm
.
cm
−
(
⎣
⎦
FIG. P36.27
f = 16.4 cm
P36.28
Let R1 = outer radius and R2 = inner radius
⎡1
1
1 ⎤
1
1
⎤ = 0.050 0 cm −1
= ( n − 1) ⎢ − ⎥ = (1.50 − 1) ⎡⎢
−
f
⎣ 2.00 m 2.50 cm ⎥⎦
⎣ R1 R2 ⎦
so
P36.29
f = 20.0 cm
For a converging lens, f is positive. We use
(a)
1 1 1
+ = .
p q f
1 1 1
1
1
1
= − =
−
=
q f p 20.0 cm 40.0 cm 40.0 cm
M =−
q = 40.0 cm
q
40.0
=−
= −1.00
p
40.0
The image is real, inverted , and located 40.0 cm past the lens.
(b)
1 1 1
1
1
= − =
−
=0
q f p 20.0 cm 20.0 cm
q = infinity
No image is formed. The rays emerging from the lens are parallel to each other.
(c)
1 1 1
1
1
1
= − =
−
=−
q f p 20.0 cm 10.0 cm
20.0 cm
q
( −20.0 )
M =− =−
= 2.00
p
10.0
q = −20.0 cm
The image is upright, virtual and 20.0 cm in front of the lens.
Image Formation
P36.30
(a)
1
1
1
+
=
32.0 cm 8.00 cm f
1 1 1
+ = :
p q f
f = 6.40 cm
so
P36.31
q
8.00 cm
=−
= −0.250
p
32.0 cm
(b)
M =−
(c)
Since f > 0, the lens is converging .
We are looking at an enlarged, upright, virtual image:
M=
h′
q
=2=−
h
p
1 1 1
+ =
p q f
q
( −2.84 cm )
=−
= +1.42 cm
2
2
so
p=−
gives
1
1
1
+
=
1.42 cm ( −2.84 cm ) f
f = 2.84 cm
FIG. P36.31
P36.32
1 1 1
+ = :
p q f
p + q = constant
We may differentiate through with respect to p:
−1 p −2 − 1q −2
−1
−1
dq
=0
dp
dq
q2
= − 2 = −M 2
dp
p
P36.33
(a)
1 1 1
+ =
p q f
1
1
1
+ =
20.0 cm q ( −32.0 cm )
1
1 ⎞
q = −⎛
+
⎝ 20.0 32.0 ⎠
so
−1
= −12.3 cm
The image is 12.3 cm to the left of the lens.
P36.34
q
( −12.3 cm )
=−
= 0.615
p
20.0 cm
(b)
M =−
(c)
See the ray diagram to the right.
The image is inverted:
M=
FIG. P36.33
h′
−1.80 m
−q
=
= −75.0 =
h 0.024 0 m
p
(a)
q + p = 3.00 m = 75.0 p + p
p = 39.5 mm
(b)
q = 2.96 m
1 1 1
1
1
= + =
+
f p q 0.039 5 m 2.96 m
q = 75.0 p
f = 39.0 mm
335
336
Chapter 36
1
1
1
1 1 1
=
we see q = −3.5 p and f = 7.50 cm for a
+ = with +
p −3.5 p 7.5 cm
p q f
converging lens.
*P36.35 Comparing
(a)
To solve, we add the fractions:
−3.5 + 1
1
=
−3.5 p
7.5 cm
3.5 p
= 7.5 cm
2.5
p = 5.36 cm
q = −3.5 ( 5.36 cm ) = −18.8 cm
(b)
M =−
−18.8 cm
q
=−
= +3.50
5.36 cm
p
(c)
I
F
O
L
F
FIG. P36.35(c)
The image is enlarged, upright, and virtual.
(d)
P36.36
In
The lens is being used as a magnifying glass. Statement: A magnifying glass with focal
length 7.50 cm is used to form an image of a stamp, enlarged 3.50 times. Find the object
distance. Locate and describe the image.
1 1 1
+ = or p −1 + q −1 = constant, we differentiate with respect to time:
p q f
−1 ( p −2 )
dp
dq
− 1 ( q −2 )
=0
dt
dt
dq − q 2 dp
= 2
dt
p dt
We must find the momentary image location q:
1
1
1
+ =
20 m q 0.3 m
q = 0.305 m
dq
( 0.305 m )
5 m s = −0.00116 m s = 1.16 mm s toward the lens .
=−
dt
( 20 m )2
2
Now
Image Formation
P36.37
(a)
q=d−p
The image distance is:
1 1 1
+ =
p q f
Thus,
337
becomes
This reduces to a quadratic equation:
1
1
1
+
=
p d−p f
p 2 + ( − d ) p + fd = 0
p=
which yields:
d ± d 2 − 4 fd d
d2
= ±
− fd
2
2
4
d
Since f < , both solutions are meaningful and the two solutions are not equal to each
4
other. Thus, there are two distinct lens positions that form an image on the screen.
(b)
The smaller solution for p gives a larger value for q, with a real, enlarged, inverted image .
The larger solution for p describes a real, diminished, inverted image .
*P36.38 (a)
1
1 1
+
=
pa qa
f
becomes
1
1
1
+ =
30.0 cm q1 14.0 cm
or
qa = 26.2 cm
⎛ −q ⎞
hb′ = hM a = h ⎜ a ⎟ = (10.0 cm ) ( −0.875) = −8.75 cm
⎝ pa ⎠
1
1
1
+
=
or
qd = 46.7 cm
20.0 cm qd 14.0 cm
hc′ = hM d = (10.0 cm ) ( −2.33) = −23.3 cm
b
c
The square is imaged as a trapezoid.
FIG. P36.38(b)
(b)
1 1
1
1 1 1
+ = becomes + =
p q 14 cm
p q f
or 1 / p = 1 / 14 cm − 1 / q
⎛ −q ⎞
⎛ 1
1⎞
h ′ = hM = h ⎜ ⎟ = (10.0 cm ) q ⎜
− ⎟
⎝ p⎠
⎝ 14 cm q ⎠
(c)
The quantity ∫
qd
qa
h ′ dq adds up the geometrical (positive) areas of thin vertical ribbons
comprising the whole area of the image. We have
46.7 cm
2
qd
qd
⎞
⎛ q − 1⎞ dq = (10.0 cm ) ⎛ q
)
=
h
dq
10
.
0
cm
−
q
(
′
⎜⎝ 28 cm
⎟⎠
∫qa
∫qa
⎝ 14 cm ⎠
26.2 cm
⎛ 46.72 − 26.22
⎞
Area = (10.0 cm ) ⎜
− 46.7 + 26.2⎟ cm = 328 cm 2
⎝
⎠
28
338
P36.39
Chapter 36
To properly focus the image of a distant object, the lens must be at a distance equal to the focal
length from the film (q1 = 65.0 mm). For the closer object:
1
1 1
+
=
p2 q2 f
becomes
1
1
1
+
=
2 000 mm q2 65.0 mm
and
⎛
2 000 ⎞
q2 = ( 65.0 mm ) ⎜
⎝ 2 000 − 65.0 ⎟⎠
The lens must be moved away from the film by a distance
⎛
2 000 ⎞
− 65.0 mm
D = q2 − q1 = ( 65.0 mm ) ⎜
m = 2.18 mm
⎝ 2 000 − 65.0 ⎟⎠
Section 36.5
P36.40
(a)
Lens Aberrations
The focal length of the lens is given by
⎛ 1
1
1⎞
1
1 ⎞
= ( n − 1) ⎜ − ⎟ = (1.53 − 1.00 ) ⎛
−
⎝
f
R
R
−
32
.
5
cm
42
.
5
cm ⎠
⎝ 1
2⎠
f = −34.7 cm
Note that R1 is negative because the center of curvature of the
first surface is on the virtual image side.
When
p=∞
the thin lens equation gives
q= f
Thus, the violet image of a very distant object is formed
q = −34.7 cm
at
The image is virtual, upright and diminshed .
FIG. P36.40
(b)
The same ray diagram and image characteristics apply for red light.
Again,
q= f
and now
1
1
1 ⎞
= (1.51 − 1.00 ) ⎛
−
⎝
f
−32.5 cm 42.5 cm ⎠
giving
f = −36.1 cm
Image Formation
P36.41
Ray h1 is undeviated at the plane surface and strikes the second
surface at angle of incidence given by
0.500 cm ⎞
h
= 1.43°
θ1 = sin −1 ⎛ 1 ⎞ = sin −1 ⎛
⎝ 20.0 cm ⎠
⎝ R⎠
0.500 ⎞
1.00 sin θ 2 = 1.60 sin θ1 = (1.60 ) ⎛
⎝ 20.0 cm ⎠
Then,
FIG. P36.41
θ 2 = 2.29°
so
The angle this emerging ray makes with the horizontal is θ 2 − θ1 = 0.860°
It crosses the axis at a point farther out by f1
0.500 cm
h1
f1 =
=
= 33.3 cm
where
tan (θ 2 − θ1 ) tan ( 0.860o)
The point of exit for this ray is distant axially from the lens vertex by
20.0 cm − ( 20.0 cm ) − ( 0.500 cm ) = 0.006 25 cm
2
2
so ray h1 crosses the axis at this distance from the vertex:
x1 = 33.3 cm − 0.006 25 cm = 33.3 cm
Now we repeat this calculation for ray h2 :
12.0 cm ⎞
θ = sin −1 ⎛
= 36.9°
⎝ 20.0 cm ⎠
12.00 ⎞
1.00 sin θ 2 = 1.60 sin θ1 = (1.60 ) ⎛
⎝ 20.0 ⎠
f2 =
θ 2 = 73.7°
12.0 cm
h2
=
= 16.0 cm
tan (θ1 − θ 2 ) tan 36.8°
(
x2 = (16.0 cm ) 20.0 cm − ( 20.0 cm ) − (12.0 cm )
P36.42
2
) = 12.0 cm
∆x = 33.3 cm − 12.0 cm = 21.3 cm
Now
Section 36.6
2
The Camera
The same light intensity is received from the subject, and the same light energy on the film is
required:
IA1 ∆t1 = IA2 ∆t 2
π d12
∆t1 =
4
2
⎛ f⎞ ⎛ 1
⎝ 4 ⎠ ⎝ 16
d2 =
π d22
∆t 2
4
1 ⎞
s
s⎞ = d22 ⎛
⎠
⎝ 128 ⎠
128 f
f
=
16 4
1.41
339
340
Chapter 36
Section 36.7
P36.43
P36.44
The Eye
1 1 1 1
1
= + = −
= −4.00 diopters = −4.00 diopters, a diverging lens
f p q ∞ 0.250 m
1 1 1
For starlight going through Nick’s glasses,
+ =
p q f
P=
1
1
1
= = −1.25 diopters
+
∞ ( −0.800 m ) f
1
1
+
= −1.25 m −1,
p ( −0.180 m )
For a nearby object,
Section 36.8
The Simple Magnifier
Section 36.9
The Compound Microscope
Section 36.10
The Telescope
P36.45
(a)
From the thin lens equation:
(b)
M =−
so p = 23.2 cm
1
1
1
=
or p = 4.17 cm
+
p ( −25.0 cm ) 5.00 cm
q
25.0 cm
25.0 cm
= 1+
= 1+
= 6.00
p
f
5.00 cm
P36.46
⎛ L ⎞ ⎛ 25.0 cm ⎞
23.0 cm ⎞ ⎛ 25.0 cm ⎞
= −⎛
Using Equation 36.26, M ≈ − ⎜ ⎟ ⎜
= −575
⎟
⎝
0.400
cm
m ⎠ ⎝ 2.50 cm ⎠
f
f
⎠
⎝ o⎠⎝
e
P36.47
fo = 20.0 m
P36.48
fe = 0.025 0 m
(a)
The angular magnification produced by this telescope is m = −
(b)
Since m < 0, the image is inverted .
(a)
The mirror-and-lens equation
1 1 1
+ =
p q f
gives
q=
Then,
M=
gives
fo
= −800 .
fe
1
1
fp
=
=
1 f − 1 p ( p − f ) fp p − f
h′
q
f
=− =−
h
p
p− f
h′ =
hf
f−p
h′ = −
hf
p
(b)
For p >> f , f − p ≈ − p. Then,
(c)
Suppose the telescope observes the space station at the zenith:
h′ = −
hf
(108.6 m ) ( 4.00 m )
=−
= −1.07 mm
m
p
407 × 10 3 m
Image Formation
P36.49
341
Let I 0 represent the intensity of the light from the nebula and θ 0 its angular diameter. With the
first telescope, the image diameter h ′ on the film is given by θ o = −
h′
as h ′ = −θ o ( 2 000 mm ).
fo
⎡ π ( 200 mm )2 ⎤
The light power captured by the telescope aperture is P1 = I 0 A1 = I 0 ⎢
⎥, and the light
4
⎣
⎦
⎡ π ( 200 mm )2 ⎤
energy focused on the film during the exposure is E1 = P1 ∆t1 = I 0 ⎢
⎥ (1.50 min ).
4
⎦
⎣
Likewise, the light power captured by the aperture of the second telescope is
⎡ π ( 60.0 mm )2 ⎤
⎡ π ( 60.0 mm )2 ⎤
and the light energy is E2 = I 0 ⎢
⎥
⎥ ∆t 2. Therefore, to
4
4
⎣
⎦
⎦
⎣
have the same light energy per unit area, it is necessary that
P2 = I 0 A2 = I 0 ⎢
2
2
I 0 ⎡⎣π ( 60.0 mm ) 4 ⎤⎦ ∆t 2 I 0 ⎡⎣π ( 200 mm ) 4 ⎤⎦ (1.50 min )
=
2
π ⎡⎣θ o ( 900 mm )2 4 ⎤⎦
π ⎡⎣θ o ( 2 000 mm ) 4 ⎤⎦
The required exposure time with the second telescope is
∆t 2 =
( 200 mm )2 ( 900 mm )2
2 (1.50 min ) = 3.38 min
( 60.0 mm )2 ( 2 000 mm )
Additional Problems
P36.50
(a)
1
1 1
1
1
= −
=
−
q1 f1 p1 5 cm 7.5 cm
∴ q1 = 15 cm
q1
15 cm
=−
= −2
p1
7.5 cm
M = M1M 2
∴ 1 = ( −2 ) M 2
q
1
∴ M2 = − = − 2
∴ p2 = 2q2
p2
2
1
1
1
1
1
1
+
=
∴
+
=
p 2 q2 f 2
2q2 q2 10 cm
m
M1 = −
∴ q2 = 15 cm, p2 = 30 cm
p1 + q1 + p2 + q2 = 7.5 cm + 15 cm + 30 cm + 15 cm = 67.5 cm
(b)
1 1
1
1
+ = =
p1′ q1′ f1 5 cm
Solve for q1′ in terms of p1′: q1′ =
5 p1′
p1′ − 5
(1)
q1′
5
, using (1).
=−
p1′
p1′ − 5
M′
q′
3
M ′ = M 1′M 2′
∴ M 2′ =
= − ( p1′ − 5) = − 2
M 1′
p2′
5
3
∴ q2′ = p2′ ( p1′ − 5)
(2)
5
1
1
1
1
and obtain p2′ in terms of p1′:
Substitute (2) into the lens equation
+
=
=
p2′ q2′ f2 10 cm
M 1′ = −
continued on next page
342
Chapter 36
p2′ =
10 ( 3 p1′ − 10 )
3 ( p1′ − 5)
(3)
Substituting (3) in (2), obtain q2′ in terms of p1′:
q2′ = 2 ( 3 p1′ − 10 )
(4)
Now, p1′ + q1′ + p2′ + q2′ = a constant.
Using (1), (3), and (4), and the value obtained in (a):
10 ( 3 p1′ − 10 )
5 p1′
+
p1′ +
+ 2 ( 3 p1′ − 10 ) = 67.5
3 ( p ′ − 5)
p1′ − 5
This reduces to the quadratic equation
21 p1′ 2 − 322.5 p1′ + 1 212.5 = 0
which has solutions p1′ = 8.784 cm and 6.573 cm.
p1′ = 8.784 cm
Case 1:
∴ p1′ − p1 = 8.784 cm − 7.5 cm = 1.28 cm
From (4):
q2′ = 32.7 cm
∴ q2′ − q2 = 32.7 cm − 15 cm = 17.7 cm
p1′ = 6.573 cm
Case 2:
∴ p1′ − p1 = 6.573 cm − 7.5 cm = −0.927 cm
From (4):
q2′ = 19.44 cm
∴ q2′ = q2 = 19.44 cm − 15 cm = 4.44 cm
From these results it is concluded that:
The lenses can be displaced in two ways. Thee first lens can be moved 1.28 cm farther from
the object and the second lens 17.7 cm toward the object. Alternatively, the firsst lens can
be moved 0.927 cm toward the objject and the second lens 4.44 cm toward thee object.
P36.51
Only a diverging lens gives an upright diminished image. The image is virtual and
q
d = p − q = p + q:
M = − so q = − Mp and d = p − Mp
p
2
d
1 1 1 1
1
− M + 1 (1 − M )
p=
+ = = +
=
=
:
1− M
p q f p − Mp
− Mp
− Md
f =
P36.52
− Md
− ( 0.500 ) ( 20.0 cm )
= −40.0 cm
2 =
(1 − M )
(1 − 0.500 )2
If M < 1, the lens is diverging and the image is virtual. d = p − q = p + q
M =−
p=
q
p
d
:
1− M
so
q = − Mp
and
d = p − Mp
1 1 1 1
1
( − M + 1) (1 − M )
=
+ = = +
=
p q f p ( − Mp )
− Mp
− Md
If M > 1, the lens is converging and the image is still virtual.
Now
d = −q − p
We obtain in this case f =
Md
( M − 1)2
.
2
f =
− Md
(1 − M )2
Image Formation
343
*P36.53 The real image formed by the concave mirror serves as a real object for the convex mirror with
p = 50 cm and q = −10 cm. Therefore,
1 1 1
= +
f p q
1
1
1
=
+
f 50 cm ( −10 cm )
gives f = −12.5 cm and R = 2 f = −25.0 cm .
*P36.54 Start with the first pass through the lens.
1
1 1
1
1
= −
=
−
q1 f1 p1 80.0 cm 100 cm
q1 = 400 cm to right of lens
For the mirror,
p2 = −300 cm
1
1
1
1
1
= −
=
−
q2 f2 p2 (−50.0 cm) (−300 cm)
q2 = −60.0 cm
For the second pass through the lens,
p3 = 160 cm
1
1 1
1
1
= −
=
−
q3 f1 p3 80.0 cm 160 cm
q3 = 160 cm to the left of lens
FIG. P36.54
1
q2
(−60.0 cm)
=−
=−
5
p2
(−300 cm)
M1 = −
q1
400 cm
=−
= −4.00
p1
100 cm
M2 = −
M3 = −
q3
160 cm
=−
= −1
160 cm
p3
M = M 1 M 2 M 3 = −0.800
Since M < 0 the final image is inverted .
*P36.55 When the meterstick coordinate of the object is 0, its object distance is pi = 32 cm. When the
meterstick coordinate of the object is x, its object distance is p = 32 cm – x. The image distance
from the lens is given by
1 1 1
+ =
p q f
1
1 1
+ =
32 − x q 26
1 32 − x − 26
=
q 26( 32 − x )
q=
832 − 26 x
6−x
The image meterstick coordinate is
x ′ = 32 + q = (32(6 − x ) + 832 − 26 x )/(6 − x ) = (1 024 cm − 58 x ) cm/(6 cm − x ) . The image
starts at the position xi = 171 cm and moves in the positive x direction, faster and faster, until it is
out at infinity when the object is at the position x = 6 cm. At this instant the rays from the top of
the object are parallel as they leave the lens. Their intersection point can be described as at x = ∞
to the right or equally well at x = −∞ on the left. From x = −∞ the image continues moving to the
right, now slowing down. It reaches, for example, −280 cm when the object is at 8 cm, and −55 cm
when the object is finally at 12 cm. The image has traveled always to the right, to infinity and
beyond.
344
P36.56
Chapter 36
1
1 1
1
1
= −
=
−
q1 f1 p1 10.0 cm 12.5 cm
so
q1 = 50.0 cm (to left of mirror)
This serves as an object for the lens (a virtual object), so
1
1
1
1
1
−
= −
=
and q2 = −50.3 cm
q2 f2 p2 ( −16.7 cm ) ( −25.0 cm )
meaning 50.3 cm to the right of the lens. Thus, the final image is located
25.3 cm to right of mirror .
M1 = −
50.0 cm
q1
=−
= −4.00
12.5 cm
p1
M2 = −
q2
( −50.3 cm )
=−
= −2.01
p2
( −25.0 cm )
M = M 1 M 2 = 8.05
Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right of
the mirror.
P36.57
A telescope with an eyepiece decreases the diameter
of a beam of parallel rays. When light is sent through
the same device in the opposite direction, the beam
expands. Send the light first through the diverging
lens. It will then be diverging from a virtual image
found like this:
1 1 1
+ =
p q f
1 1
1
+ =
∞ q −12 cm
FIG. P36.57
q = −12 cm
Use this image as a real object for the converging lens, placing it at the focal point on the object
side of the lens, at p = 21cm. Then
1 1 1
+ =
p q f
1
1
1
+ =
21 cm q 21 cm
q=∞
The exiting rays will be parallel. The lenses must be 21.0 cm − 12.0 cm = 9.00 cm apart.
By similar triangles,
d2 21 cm
=
= 1.75 times
d1 12 cm
Image Formation
*P36.58 (a)
345
For the lens in air,
⎛ 1
1
1⎞
= ( n − 1) ⎜ − ⎟
f
⎝ R1 R2 ⎠
⎛ 1
1
1⎞
= (1.55 − 1) ⎜ − ⎟
79 cm
R
R
⎝ 1
2⎠
For the same lens in water
⎞⎛ 1
1 ⎛ n2
1⎞
= ⎜ − 1⎟ ⎜ − ⎟
f ′ ⎝ n1
⎠ ⎝ R1 R2 ⎠
1 ⎛ 1.55 ⎞ ⎛ 1
1⎞
=
−1 ⎜ − ⎟
f ′ ⎝ 1.33 ⎠ ⎝ R1 R2 ⎠
By division,
1 79 cm
0.55
f′
=
=
1/ f ′
0.165 79 cm
f ′ = 79 cm ( 3.32 ) = 263 cm
(b)
P36.59
The path of a reflected ray does not depend on the refractive index of the medium which the
reflecting surface bounds. Therefore the focal length of a mirror does not change when it is
R
put into a different medium: f ′ = = f = 79.0 cm .
2
A hemisphere is too thick to be described as a thin lens.
The light is undeviated on entry into the flat face. We next
consider the light’s exit from the second surface, for which
R = −6.00 cm.
p=∞
The incident rays are parallel, so
Then,
n1 n2 n2 − n1
+
=
p q
R
becomes
0+
1 1.00 − 1.56
=
q −6.00 cm
q = 10.7 cm
and
P36.60
FIG. P36.59
P
I=
(b)
I=
(c)
1 1 1
+ = :
p q f
1
1
1
+ =
7.20 m q 0.350 m
so
q = 0.368 m
and
M=
4π r 2
P
4π r 2
=
4.50 W
(a)
=
4π (1.60 × 10
−2
m)
2
= 1.40 kW m 2
4.50 W
= 6.91 mW m 2
4π ( 7.20 m )2
h′
q
0.368 m
=− =−
3.20 cm
p
7.20 m
h ′ = 0.164 cm
(d)
The lens intercepts power given by
and puts it all onto the image where
π
P = IA = ( 6.91 × 10 −3 W m 2 ) ⎡⎢ ( 0.150 m )2 ⎤⎥
⎣4
⎦
2
−3
2
P ( 6.91 × 10 W m ) ⎡⎣π (15.0 cm ) 4 ⎤⎦
I= =
A
π ( 0.164 cm )2 4
I = 58.1 W m 2
346
P36.61
Chapter 36
f1 p1
( −6.00 cm ) (12.0 cm )
= −4.00 cm
=
p1 − f1 12.0 cm − ( −6.00 cm )
From the thin lens equation,
q1 =
When we require that q2 → ∞,
the thin lens equation becomes p2 = f2.
In this case,
p2 = d − ( −4.00 cm )
Therefore,
d + 4.00 cm = f2 = 12.0 cm
and
d = 8.00 cm
FIG. P36.61
P36.62
(a)
For the light the mirror intercepts,
P = I 0 A = I 0π Ra2
350 W = (1 000 W m 2 ) π Ra2
(b)
and
Ra = 0.334 m or larger
In
1 1 1 2
+ = =
p q f R
we have
p→∞
so
q=
so
⎛ h⎞
R ⎡
π rad ⎞ ⎤
R
h ′ = − q ⎜ ⎟ = − ⎛ ⎞ ⎢ 0.533° ⎛
= − ⎛ ⎞ ( 9.30 m rad )
⎥
⎝
⎝
⎠
⎝
⎠
⎝ p⎠
2 ⎣
180 ° ⎦
2⎠
where
then
R
2
h′
q
M = =−
h
p
h
is the angle the Sun subtends. The intensity at the image is
p
I=
P
π h′ 2 4
=
4 I 0π Ra2
4 I 0 Ra2
=
2
π h′ 2
( R 2)2 ( 9.30 × 10 −3 raad )
120 × 10 3 W m 2 =
so
16 (1 000 W m 2 ) Ra2
R 2 ( 9.30 × 10 −3 rad )
Ra
= 0.025 5 or larger
R
2
Image Formation
P36.63
347
R
= +1.50 m. In addition, because the distance to the Sun is so much larger
2
than any other distances, we can take p = ∞.
For the mirror, f =
The mirror equation,
1 1 1
+ = , then gives
p q f
q = f = 1.50 m
q h′
=
p h
h
the magnification is nearly zero, but we can be more precise: is the angular diameter of the
p
object. Thus, the image diameter is
M =−
Now, in
h′ = −
P36.64
(a)
⎛ 1
1
1⎞
= ( n − 1) ⎜ + ⎟
f
⎝ R1 R2 ⎠
The lens makers’ equation,
becomes:
(b)
π rad ⎞
hq
= ( −0.533o) ⎛
(1.50 m ) = −0.014 0 m = −1.40 cm
⎝ 180 ° ⎠
p
⎡
⎤
1
1
1
−
= ( n − 1) ⎢
⎥ giving n = 1.99 .
5.00 cm
⎣ 9.00 cm ( −11.0 cm ) ⎦
As the light passes through the lens for the first time, the thin lens equation
1
1
1
1 1 1
+ =
+ =
becomes:
8.00 cm q1 5.00 cm
p1 q1 f
or
q1 = 13.3 cm,
and
M1 = −
q1
13.3 cm
=−
= −1.67
p1
8.00 cm
This image becomes the object for the concave mirror with:
pm = 20.0 cm − q1 = 20.0 cm − 13.3 cm = 6.67 cm
and
The mirror equation becomes:
1
1
1
+
=
6.67 cm qm 4.00 cm
giving
qm = 10.0 cm
and
f =
M2 = −
R
= +4.00 cm
2
10.0 cm
qm
=−
= −1.50
6.67 cm
pm
The image formed by the mirror serves as a real object for the lens on the second pass of
the light through the lens with:
p3 = 20.0 cm − qm = +10.0 cm
The thin lens equation yields:
1
1
1
+ =
10.0 cm q3 5.00 cm
or
q3 = 10.0 cm
The final image is a real image located
The overall magnification is
(c)
and
M3 = −
q3
10.0 cm
=−
= −1.00
10.0 cm
p3
10.0 cm to the left of the lens .
M total = M 1 M 2 M 3 = −2.50
Since the total magnification is negative, this final image is inverted .
348
P36.65
Chapter 36
In the original situation,
p1 + q1 = 1.50 m
In the final situation,
p2 = p1 + 0.900 m
and
q2 = q1 − 0.900 m = 0.600 m − p1
Our lens equation is
1 1 1
1
1
+ = =
+
p1 q1 f p2 q2
FIG. P36.65
Substituting, we have
1
1
1
1
+
=
+
p1 1.50 m − p1 p1 + 0.900 0.600 − p1
Adding the fractions,
1.50 m − p1 + p1
0.600 − p1 + p1 + 0.900
=
p1 (1.50 m − p1 ) ( p1 + 0.900 ) ( 0.600 − p1 )
Simplified, this becomes
p1 (1.50 m − p1 ) = ( p1 + 0.900 ) ( 0.600 − p1 )
(a)
Thus,
p1 =
(b)
1
1
1
=
+
f 0.300 m 1.50 m − 0.300 m
(c)
The second image is real, inverted, and diminished
M =−
with
P36.66
0.540
m = 0.300 m
1.80
and
p2 = p1 + 0.900 = 1.20 m
f = 0.240 m
q2
= −0.250
p2
The object is located at the focal point of the upper mirror. Thus,
the upper mirror creates an image at infinity (i.e., parallel rays
leave this mirror).
The lower mirror focuses these parallel rays at its focal point,
located at the hole in the upper mirror.
Thus, the
image is real, inverted, and actual size .
For the upper mirror:
1 1 1
+ = :
p q f
1
1
1
+ =
7.50 cm q1 7.50 cm
q1 = ∞
For the lower mirror:
1 1
1
+
=
∞ q2 7.50 cm
q2 = 7.50 cm
Light directed into the hole in the upper mirror reflects as shown,
to behave as if it were reflecting from the hole.
FIG. P36.66
Image Formation
P36.67
(a)
For lens one, as shown in the first figure,
1
1
1
+ =
40.0 cm q1 30.0 cm
q1 = 120 cm
M1 = −
q1
120 cm
=−
= −3.00
p1
40.0 cm
This real image I1 = O2 is a virtual object
for the second lens. That is, it is behind
the lens, as shown in the second figure.
The object distance is
p2 = 110 cm − 120 cm = −10.0 cm
1
1
1
:
+
=
−10.0 cm q2 −20.0 cm
q2 = 20.0 cm
M2 = −
20.0 cm
q2
= +2.00
=−
p2
( −10.0 cm )
M overall = M 1 M 2 = −6.00
(b)
M overall < 0, so final image
is inverted .
(c)
If lens two is a converging lens (third figure):
1
1
1
+
=
−10.0 cm q2 20.0 cm
q2 = 6.67 cm
M2 = −
6.67 cm
( −10.0 cm )
= +0.667
M overall = M 1 M 2 = −2.00
Again, M overall < 0 and the final image is inverted .
FIG. P36.67
349
350
P36.68
Chapter 36
The first lens has focal length described by
⎛ 1
1
1 ⎞
1 1
n −1
−
= ( n1 − 1) ⎛ − ⎞ = − 1
= ( n1 − 1) ⎜
⎝ ∞ R⎠
f1
R
⎝ R11 R12 ⎟⎠
For the second lens
2 ( n2 − 1)
⎛ 1
1
1 ⎞
1
1 ⎞
−
= ( n2 − 1) ⎛
=+
= ( n2 − 1) ⎜
−
⎟
⎝
⎠
f2
+R −R
R
⎝ R21 R22 ⎠
Let an object be placed at any distance p1 large compared to the thickness of the doublet. The first
lens forms an image according to
1 1
1
+ =
p1 q1 f1
1 − n1 + 1 1
=
−
q1
R
p1
This virtual ( q1 < 0 ) image is a real object for the second lens at distance p2 = − q1. For the
second lens
1
1
1
+
=
p2 q2 f2
1 2n2 − 2 1 2n2 − 2 1 2n2 − 2 − n1 + 1 1 2n2 − n1 − 1 1
=
−
=
−
=
+ =
+
−
q2
R
p2
R
q1
R
R
p1
R
p1
Then
1
1 2n2 − n1 − 1
1 2n − n1 − 1
so the doublet behaves like a single lens with = 2
.
+
=
p1 q2
R
f
R
ANSWERS TO EVEN PROBLEMS
P36.2
4.58 m
P36.4
10.0 ft, 30.0 ft, 40.0 ft
P36.6
(a) 13.3 cm, –0.333, real and inverted
formed.
P36.8
at q = −0.267 m virtual, upright, and diminished with M = 0.026 7
P36.10
at 3.33 m from the deepest point of the niche
P36.12
At 0.708 cm in front of the reflecting surface. Image is virtual, upright, and diminished.
P36.14
7.90 mm
P36.16
(a) 15.0 cm
(b) 60.0 cm
P36.18
(a) 25.6 m
(b) 0.058 7 rad
(c) 2.51 m
P36.20
4.82 cm
P36.22
(a) 45.0 cm
(b) –90.0 cm
(c) –6.00 cm
(b) 20.0 cm, –1.00, real and inverted
(d) 0.023 9 rad
(c) No image is
(e) 62.8 m from your eyes
Image Formation
351
P36.24
See the solution.
P36.26
1.50 cm s
P36.28
20.0 cm
P36.30
(a) 6.40 cm
P36.32
See the solution.
P36.34
(a) 39.0 mm
P36.36
1.16 mm s toward the lens
P36.38
(a) qa = 26.2 cm qd = 46.7 cm hb = −8.75 cm hc = −23.3 cm. See the solution. (b) The equation follows from h/h = −q/p and 1/p + 1/q = 1/f. (c) The integral stated adds up the areas of
ribbons covering the whole image, each with vertical dimension | h | and horizontal width dq.
328 cm2
P36.40
(a) at q = −34.7 cm virtual, upright, and diminished
diminished
P36.42
f
1.41
P36.44
23.2 cm
P36.46
−575
P36.48
(a) See the solution.
P36.50
(a) 67.5 cm (b) The lenses can be displaced in two ways. The first lens can be displaced 1.28 cm
farther away from the object, and the second lens 17.7 cm toward the object. Alternatively, the first
lens can be displaced 0.927 cm toward the object and the second lens 4.44 cm toward the object.
P36.52
if M < 1, f =
P36.54
160 cm to the left of the lens, inverted, M = −0.800
P36.56
25.3 cm to right of mirror, virtual, upright, enlarged 8.05 times
P36.58
(a) 263 cm
P36.60
(a) 1.40 kW m 2 (b) 6.91 mW m 2
P36.62
(a) 0.334 m or larger
P36.64
(a) 1.99
P36.66
See the solution; real, inverted, and actual size.
P36.68
See the solution.
(b) −0.250 (c) converging
(b) 39.5 mm
(b) h ′ = −
− Md
,
(1 − M )2
hf
p
(b) at q = −36.1 cm virtual, upright, and
(c) −1.07 mm
if M > 1, f =
Md
( M − 1)2
(b) 79.0 cm
(b)
(c) 0.164 cm
(d) 58.1 W m 2
Ra
= 0.025 5 or larger
R
(b) 10.0 cm to the left of the lens; 2.50
(c) inverted
37
Interference of Light Waves
CHAPTER OUTLINE
37.1
37.2
37.3
37.4
37.5
37.6
37.7
Conditions for Interference
Young’s Double-Slit Experiment
Light Waves in Interference
Intensity Distribution of the
Double-Slit Interference Pattern
Change of Phase Due to
Reflection
Interference in Thin Films
The Michelson Interferometer
ANSWERS TO QUESTIONS
Q37.1
Q37.2
(a)
Two waves interfere constructively if their
path difference is zero, or an integral multiple
of the wavelength, according to δ = mλ , with
m = 0, 1, 2, 3, . . . .
(b)
Two waves interfere destructively if their path
difference is a half wavelength, or an odd
λ
1
multiple of , described by δ = ⎛ m + ⎞ λ , with
⎝
2
2⎠
m = 0, 1, 2, 3, . . . .
The light from the flashlights consists of many different
wavelengths (that’s why it’s white) with random time
differences between the light waves. There is no
coherence between the two sources. The light from
the two flashlights does not maintain a constant phase
relationship over time. These three equivalent statements
mean no possibility of an interference pattern.
*Q37.3 (i) The angles in the interference pattern are controlled by λ /d, which we estimate in each case:
(a) 0.45 µm/400 µm ≈ 1.1 × 10−3 (b) 0.7 µm/400 µm ≈ 1.6 × 10−3 (c) and (d) 0.7 µm/800 µm ≈
0.9 × 10−3. The ranking is b > a > c = d.
(ii) Now we consider Lλ /d: (a) 4 m (0.45 µm/400 µm) ≈ 4.4 mm (b) 4 m (0.7 µm/400 µm) ≈
7 mm (c) 4 m(0.7 µm/800 µm) ≈ 3 mm (d) 8 m(0.7 µm/800 µm) ≈ 7 mm. The ranking is
b = d > a > c.
*Q37.4 Yes. A single beam of laser light going into the slits divides up into several fuzzy-edged beams
diverging from the point halfway between the slits.
λair
. Since
nwater
the angles between positions of light and dark bands are proportional to λ , the underwater fringe
separations decrease.
*Q37.5 Answer (c). Underwater, the wavelength of the light decreases according to λwater =
Q37.6
Every color produces its own pattern, with a spacing between the maxima that is characteristic
of the wavelength. With white light, the central maximum is white. The first side maximum is
a full spectrum with violet on the inner edge and red on the outer edge on each side. Each side
maximum farther out is in principle a full spectrum, but they overlap one another and are hard to
distinguish. Using monochromatic light can eliminate this problem.
*Q37.7 With two fine slits separated by a distance d slightly less than λ , the equation d sin θ = 0 has
the usual solution θ = 0. But d sin θ = 1λ has no solution. There is no first side maximum.
d sin θ = (1/2)λ has a solution. Minima flank the central maximum on each side. Answer (b).
353
354
Q37.8
Chapter 37
As water evaporates from the “soap” bubble, the thickness of the bubble wall approaches
zero. Since light reflecting from the front of the water surface is phase-shifted 180° and light
reflecting from the back of the soap film is phase-shifted 0°, the reflected light meets the
conditions for a minimum. Thus the soap film appears black, as in the textbook illustration
accompanying this question.
*Q37.9 Answer (b). If the thickness of the oil film were smaller than half of the wavelengths of visible
light, no colors would appear. If the thickness of the oil film were much larger, the colors would
overlap to mix to white or gray.
*Q37.10 (i) Answer (b). If the oil film is brightest where it is thinnest, then nair < noil < nglass . With this
condition, light reflecting from both the top and the bottom surface of the oil film will undergo
phase reversal. Then these two beams will be in phase with each other where the film is very thin.
This is the condition for constructive interference as the thickness of the oil film decreases toward
zero. If the oil film is dark where it is thinnest, then nair < noil > nglass. In this case, reflecting light
undergoes phase reversal upon reflection from the front surface but no phase reversal upon
reflection from the back surface. The two reflected beams are 180° out of phase and interfere
destructively as the oil film thickness goes to zero.
(ii) Yes. It should be lower in index than both kinds of glass.
(iii) Yes. It should be higher in refractive index than both kinds of glass.
(iv) No.
Q37.11
If R is large, light reflecting from the lower surface of the lens can interfere with light reflecting from the upper surface of the flat. The latter undergoes phase reversal on reflection while
the former does not. Where there is negligible distance between the surfaces, at the center of the
pattern you will see a dark spot because of the destructive interference associated with the 180°
phase shift. Colored rings surround the dark spot. If the lens is a perfect sphere the rings are
perfect circles. Distorted rings reveal bumps or hollows on the fine scale of the wavelength of
visible light.
Q37.12 A camera lens will have more than one element, to correct (at least) for chromatic aberration. It
will have several surfaces, each of which would reflect some fraction of the incident light. To
maximize light throughput the surfaces need antireflective coatings. The coating thickness is
chosen to produce destructive interference for reflected light of some wavelength.
*Q37.13 (i) Answer (c). The distance between nodes is one-half the wavelength.
(ii) Answer (d). Moving one mirror by 125 nm lengthens the path of light reflecting from it by
250 nm. Since this is half a wavelength, the action reverses constructive into destructive
interference.
(iii) Answer (e). The wavelength of the light in the film is 500 nm /2 = 250 nm. If the film is made
62.5 nm thicker, the light reflecting inside the film has a path length 125 nm greater. This is half a
wavelength, to reverse constructive into destructive interference.
*Q37.14 Answer (a). If the mirrors do not move the character of the interference stays the same. The light
does not get tired before entering the interferometer or undergo any change on the way from the
source to the half-silvered mirror.
Interference of Light Waves
355
SOLUTIONS TO PROBLEMS
Section 37.1
Conditions for Interference
Section 37.2
Young’s Double-Slit Experiment
Section 37.3
Light Waves in Interference
P37.1
∆ybright =
P37.2
ybright =
−9
λ L ( 632.8 × 10 ) ( 5.00 )
=
m = 1.58 cm
d
2.00 × 10 −4
λL
m
d
λ=
For m = 1,
−3
−4
yd ( 3.40 × 10 m ) ( 5.00 × 10 m )
=
= 515 nm
L
3.30 m
P37.3
400 m
300 m
1 000 m
FIG. P37.3
Note, with the conditions given, the small angle approximation does not work well. That is, sin θ ,
tan θ , and θ are significantly different. We treat the interference as a Fraunhofer pattern.
(a)
At the m = 2 maximum, tan θ =
400 m
= 0.400
1 000 m
θ = 21.8°
So
(b)
λ=
d sin θ ( 300 m ) sin 21.8°
=
= 55.7 m
m
2
The next minimum encountered is the m = 2 minimum, and at that point,
1
d sin θ = ⎛ m + ⎞ λ
⎝
2⎠
5
λ
2
which becomes
d sin θ =
or
sin θ =
and
θ = 27.7°
so
y = (1 000 m ) tan 27.7° = 524 m
5 λ 5 ⎛ 55.7 m ⎞
=
= 0.464
2 d 2 ⎝ 300 m ⎠
Therefore, the car must travel an additional 124 m .
If we considered Fresnel interference, we would more precisely find
(a) λ =
1
2
(
)
550 2 + 1 000 2 − 250 2 + 1 000 2 = 55.2 m and (b) 123 m
356
P37.4
Chapter 37
λ=
v 354 m s
=
= 0.177 m
f 2 000 s −1
(a)
d sin θ = mλ
so
( 0.300 m ) sin θ = 1( 0.177 m )
and
θ = 36.2°
(b)
d sin θ = mλ
so
d sin 36.2° = 1 ( 0.030 0 m )
and
d = 5.08 cm
(c)
(1.00 × 10
so
λ = 590 nm
f =
P37.5
−6
m ) sin 36.2° = (1) λ
c 3.00 × 108 m s
=
= 508 THz
λ
5.90 × 10 −7 m
y
1
d sin θ = ⎛ m + ⎞ λ
⎝
2⎠
The first minimum is described by
m=0
In the equation
λ⎛
1
9+ ⎞
⎝
d
2⎠
y
tan θ =
L
sin θ ≈ tan θ
9.5λ 9.5λ L
d=
=
sin θ
y
Also,
but for small θ ,
Thus,
d=
L
sin θ =
and the tenth by m = 9:
9.5 ( 5 890 × 10 −10 m ) ( 2.00 m )
7.26 × 10 −3 m
= 1.54 × 10
−3
d
Source
FIG. P37.5
m = 1.54 mm
*P37.6
Problem statement: A single oscillator makes the two speakers of a boom box, 35.0 cm apart,
vibrate in phase at 1.62 kHz. At what angles, measured from the perpendicular bisector of the line
joining the speakers, will a distant observer hear maximum sound intensity? Minimum sound
intensity? The ambient temperature is 20°C.
We solve the first equation for λ , substitute into the others, and solve for each angle to find this
answer: The wavelength of the sound is 21.2 cm. Interference maxima occur at angles of 0° and
37.2° to the left and right. Minima occur at angles of 17.6° and 65.1°. No second-order or higherorder maximum exists. No angle exists, smaller or larger than 90°, for which sin θ 2 loud = 1.21. No
location exists in the Universe that is two wavelengths farther from one speaker than from the other.
P37.7
(a)
ybright =
y=
(b)
y
For the bright fringe,
mλ L
where m = 1
d
(546.1 × 10
−9
L = 1.20 m
m ) (1.20 m )
0.250 × 10 −3 m
For the dark bands, ydark =
y2 − y1 =
=
= 2.62 × 10 −3 m = 2.62 mm
Source
λL ⎛
1
m + ⎞ ; m = 0, 1, 2, 3, . . .
⎝
d
2⎠
d = 0.250 m
bright bright bright
dark
dark
dark
λ L ⎡⎛ 1 ⎞ ⎛
1 ⎤ λL
1+
− 0+ ⎞⎥ =
(1)
d ⎢⎣⎝
2⎠ ⎝
2⎠ ⎦ d
(546.1 × 10
−9
m ) (1.20 m )
0.250 × 10 −3 m
∆y = 2.62 mm
y1
y2
FIG. P37.7
Interference of Light Waves
P37.8
357
Location of A = central maximum,
Location of B = first minimum.
P37.9
So,
∆y = [ ymin − ymax ] =
Thus,
d=
1
1 λL
λL ⎛
0+ ⎞ −0=
= 20.0 m
d ⎝
2⎠
2 d
λL
( 3.00 m ) (150 m )
=
= 11.3 m
2 ( 20.0 m )
40.0 m
Taking m = 0 and y = 0.200 mm in Equations 37.3 and 37.4 gives
L≈
−3
−3
2dy 2 ( 0.400 × 10 m ) ( 0.200 × 10 m )
=
= 0.362 m
λ
442 × 10 −9 m
L ≈ 36.2 cm
Bright
Dark
0.2 mm
Bright
0.2 mm
Geometric optics or a particle theory of light would incorrectly
predict bright regions opposite the slits and darkness in between.
But, as this example shows, interference can produce just the opposite.
Dark
L
Bright
FIG. P37.9
P37.10
d sin θ = mλ
At 30.0°,
(3.20 × 10
−4
m ) sin 30.0° = m ( 500 × 10 −9 m )
so
m = 320
There are 320 maxima to the right, 320 to the left, and one for m = 0 straight ahead.
There are 641 maxima .
P37.11
P37.12
Observe that the pilot must not only home in on the airport, but must be headed in the right
direction when she arrives at the end of the runway.
c 3 × 108 m s
=
= 10.0 m
f 30 × 10 6 s −1
(a)
λ=
(b)
The first side maximum is at an angle given by d sin θ = (1) λ .
y
θ = 14.5°
tan θ =
( 40 m ) sin θ = 10 m
L
y = L tan θ = ( 2 000 m ) tan 14.5° = 516 m
(c)
The signal of 10-m wavelength in parts (a) and (b) would show maxima at 0°, 14.5°, 30.0°,
48.6°, and 90°. A signal of wavelength 11.23 m would show maxima at 0°, 16.3°, 34.2°,
and 57.3°. The only value in common is 0°. If λ1 and λ2 were related by a ratio of small
λ
n
integers (a just musical consonance!) in 1 = 1 , then the equations d sin θ = n2 λ1 and
λ2 n2
d sin θ = n1λ2 would both be satisfied for the same nonzero angle. The pilot could come flying
in with that inappropriate bearing, and run off the runway immediately after touchdown.
In d sin θ = mλ
dy mλ dL
=
=
dt
d dt
y
= mλ
L
1 ( 633 × 10 −9 m )
d
( 0.3 × 10
−3
m)
y=
mλ L
d
3 m s = 6.33 mm s
358
P37.13
Chapter 37
φ=
2π
2π ⎛ y ⎞
d sin θ =
d
λ
λ ⎝ L⎠
2π
(a)
φ=
(b)
φ=
(c)
If φ = 0.333 rad =
(5.00 × 10
−7
m)
(1.20 × 10
−4
m ) sin (0.500°) = 13.2 rad
⎛ 5.00 × 10 −3 m ⎞
2π
−4
1
20
10
.
m
×
= 6.28 rad
(
)
⎜⎝
1.220 m ⎟⎠
(5.00 × 10 −7 m )
2π d sin θ
λ
⎡ 5.00 × 10 −7 m ) ( 0.333 rad ) ⎤
⎛ λφ ⎞
−1 (
=
θ = sin −1 ⎜
sin
⎥
⎢
⎝ 2π d ⎟⎠
2π (1.20 × 10 −4 m )
⎥⎦
⎢⎣
θ = 1.27 × 10 −2 deg
(d)
If d sin θ =
⎡ 5 × 10 −7 m ⎤
⎛ λ⎞
θ = sin −1 ⎜ ⎟ = sin −1 ⎢
⎥
−4
⎝ 4d ⎠
⎢⎣ 4 (1.20 × 10 m ) ⎥⎦
λ
4
θ = 5.97 × 10 −2 deg
P37.14
The path difference between rays 1 and 2 is:
δ = d sin θ1 − d sin θ 2
For constructive interference, this path difference must be equal to an integral number of
wavelengths: d sin θ1 − d sin θ 2 = mλ , or
d ( sin θ1 − sin θ 2 ) = mλ
*P37.15 (a)
The path difference δ = d sin θ and when L >> y
δ=
(b)
(c)
Section 37.4
P37.16
(a)
(b)
−2
−4
yd (1.80 × 10 m ) (1.50 × 10 m )
=
= 1.93 × 10 −6 m = 1.93 µ m
L
1.40 m
δ 1.93 × 10 −6 m
=
= 3.00, or δ = 3.00 λ
λ 6.43 × 10 −7 m
Point P will be a maximum because the path difference is an integer multiple of the
wavelength.
Intensity Distribution of the Double-Slit Interference Pattern
I
φ
= cos 2 ⎛ ⎞
⎝ 2⎠
I max
(Equation 37.12)
Therefore,
φ = 2 cos −1
δ=
I
I max
= 2 cos −1 0.640 = 1.29 rad
λφ ( 486.1 nm ) (1.29 rad )
=
= 99.6 nm
2π
2π
Interference of Light Waves
P37.17
I av = I max cos 2 ⎛
⎝
For small θ ,
π d sin θ ⎞
λ ⎠
y
sin θ =
L
I av = 0.750 I max
and
y=
I av
λL
cos −1
I max
πd
y=
(6.00 × 10 ) (1.20 m ) cos
π ( 2.50 × 10 m )
−7
P37.18
359
−1
−3
0.750 I max
= 48.0 µ m
I max
π yd ⎞
I = I max cos 2 ⎛
⎝ λL ⎠
I
I max
⎡ π ( 6.00 × 10 −3 m ) (1.80 × 10 −4 m ) ⎤
= cos 2 ⎢
⎥ = 0.968
−9
⎢⎣ ( 656.3 × 10 m ) ( 0.800 m ) ⎥⎦
*P37.19 We use trigonometric identities to write
6 sin(100pt) + 8 sin(100pt + p/2) = 6 sin(100pt) + 8 sin(100pt) cos(p/2) + 8 cos(100pt)sin(p/2)
E1 + E2 = 6 sin(100p t) + 8 cos(100p t)
and ER sin(100p t + φ ) = ER sin(100p t) cos φ + ER cos (100p t) sin φ
The equation E1 + E2 = ER sin(100p t + φ ) is satisfied if we require just
6 = ER cos φ and 8 = ER sin φ
or 62 + 82 = ER (cos2 φ + sin2 φ )
ER = 10
and tan φ = sin φ /cos φ = 8/6 = 1.33
φ = 53.1°
2
π d sin θ ⎞
*P37.20 In I av = I max cos 2 ⎛
for angles between −0.3°
⎝
λ ⎠
and +0.3° we take sin θ = θ to find
⎛ π 250 µ m θ ⎞
I = I max cos 2 ⎜
⎝ 0.5461 µ m ⎟⎠
IⲐImax
I / I max = cos 2 (1438 θ )
−0.2
0
0.2
q , degrees
This equation is correct assuming θ is in radians; but
we can then equally well substitute in values for θ in
FIG. P37.20
degrees and interpret the argument of the cosine function
as a number of degrees. We get the same answers for θ negative and for θ positive. We evaluate
θ , degrees −0.3 −0.25 −0.2 −0.15 −0.1 −0.05 0 0.05
0.1 0.15
0.2 0.25
0.3
I/Imax
0.101 1.00 0.092 0.659 0.652 0.096 1 0.096 0.652 0.659 0.092 1.00 0.101
The cosine-squared function has maximum values of 1 at θ = 0, at 1438 θ = 180° with θ = 0.125°,
and at 1438 θ = 360° with θ = 0.250°. It has minimum values of zero halfway between the maximum
values. The graph then has the appearance shown.
360
P37.21
Chapter 37
(a)
From Equation 37.9,
2π d
2π d
y
φ=
⋅ 2
sin θ =
λ
λ
y + D2
φ≈
(b)
I
I max
−3
−3
2π yd 2π ( 0.850 × 10 m ) ( 2.50 × 10 m )
= 7.95 rad
=
λD
(600 × 10 −9 m ) ( 2.80 m )
=
cos 2 ⎡⎣(π d λ ) sin θ ⎤⎦
cos ⎡⎣(π d λ ) sin θ max ⎤⎦
2
=
cos 2 (φ 2 )
cos 2 mπ
I
φ
7.95 rad ⎞
= cos 2 = cos 2 ⎛
= 0.453
⎝
I max
2
2 ⎠
*P37.22 (a)
The resultant amplitude is
2π
d sin θ
λ
Er = E0 ( sin ω t + sin ω t cos φ + cos ω t sin φ + sin ω t cos 2φ + cos ω t sin 2φ )
Er = E0 sin ω t + E0 sin (ω t + φ ) + E0 sin (ω t + 2φ ), where φ =
Er = E0 ( sin ω t ) (1 + cos φ + 2 cos 2 φ − 1) + E0 ( cos ω t ) ( sin φ + 2 sin φ cos φ )
Er = E0 (1 + 2 cos φ ) ( sin ω t cos φ + cos ω t sin φ ) = E0 (1 + 2 cos φ ) sin (ω t + φ )
2 1
I ∝ Er2 = E02 (1 + 2 cos φ ) ⎛ ⎞
⎝ 2⎠
1
2
where the time average of sin (ω t + φ ) is .
2
1
From one slit alone we would get intensity I max ∝ E02 ⎛ ⎞ so
⎝ 2⎠
Then the intensity is
2π d sin θ ⎞ ⎤
⎡
I = I max ⎢1 + 2 cos ⎛
⎝
⎠ ⎥⎦
λ
⎣
(b)
2
Look at the N = 3 graph in the textbook Figure 37.8. Minimum intensity is zero, attained
1
where cos φ = − . One relative maximum occurs at cos φ = −1.00, where
2
2
I = I max [1 − 2] = I max
The larger local maximum happens where cos φ = +1.00, giving I = I max [1 + 2]2 = 9.00 I max .
The ratio of intensities at primary versus secondary maxima is 9.00 .
Interference of Light Waves
Section 37.5
Change of Phase Due to Reflection
Section 37.6
Interference in Thin Films
P37.23
(a)
361
The light reflected from the top of the oil film undergoes phase
reversal. Since 1.45 > 1.33, the light reflected from the bottom
undergoes no reversal. For constructive interference of reflected
light, we then have
or
1
2nt = ⎛ m + ⎞ λ
⎝
2⎠
2nt
2 (1.45) ( 280 nm )
λm =
=
m + (1 2 )
m + (1 2 )
Substituting for m gives:
m = 0,
λ0 = 1 620 nm (infrared)
m = 1,
λ1 = 541 nm (green)
m = 2,
λ2 = 325 nm (ultraviolet)
FIG. P37.23
Both infrared and ultraviolet light are invisible to the human eye, so the dominant color in
reflected light is green .
(b)
The dominant wavelengths in the transmitted light are those that produce destructive
interference in the reflected light. The condition for destructive interference upon
reflection is
2nt = mλ
or
λm =
2nt 812 nm
=
m
m
Substituting for m gives:
m = 1,
λ1 = 812 nm (near infrared)
m = 2,
λ2 = 406 nm (violet)
m = 3,
λ3 = 271 nm (ultraviolet)
Of these, the only wavelength visible to the human eye (and hence the dominant
wavelength observed in the transmitted light) is 406 nm. Thus, the dominant color
in the transmitted light is violet .
P37.24
Light reflecting from the first surface suffers phase reversal. Light reflecting from the second
surface does not, but passes twice through the thickness t of the film. So, for constructive
interference, we require
where
Then
λn
+ 2t = λn
2
λ
λn = is the wavelength in the material.
n
λ
λ
2t = n =
2 2n
λ = 4 nt = 4 (1.33) (115 nm ) = 612 nm
362
P37.25
Chapter 37
Since 1 < 1.25 < 1.33, light reflected both from the top and from the bottom surface of the oil
suffers phase reversal.
mλcons
For constructive interference we require
2t =
n
and for destructive interference,
Then
t=
Therefore,
P37.26
⎡ m + (1 2 ) ⎤⎦ λdes
2t = ⎣
n
λcons
1
640 nm
= 1+
=
= 1.25 and m = 2
λdest
2 m 512 nm
2 ( 640 nm )
= 512 nm
2 (1.25)
Treating the anti-reflectance coating like a camera-lens coating,
1 λ
2t = ⎛ m + ⎞
⎝
2⎠ n
λ 3.00 cm
Let m = 0:
t=
=
= 0.500 cm
4 n 4 (1.50 )
This anti-reflectance coating could be easily countered by changing the wavelength of the radar
to 1.50 cm. Then the coating would exhibit maximum reflection!
P37.27
1
2nt = ⎛ m + ⎞ λ
⎝
2⎠
Minimum
P37.28
so
1 λ
t = ⎛m + ⎞
⎝
2 ⎠ 2n
1 ( 500 nm )
t=⎛ ⎞
= 96.2 nm
⎝ 2 ⎠ 2 (1.30 )
Since the light undergoes a 180° phase change at each surface of the film, the condition for
2nt
constructive interference is 2nt = mλ , or λ =
. The film thickness is
m
−5
−7
t = 1.00 × 10 cm = 1.00 × 10 m = 100 nm. Therefore, the wavelengths intensified in the
reflected light are
2 (1.38 ) (100 nm ) 276 nm
=
where m = 1, 2, 3, . . .
m
m
or λ1 = 276 nm, λ2 = 138 nm, . . . . All reflection maxima are in the ultraviolet and beyond.
λ=
No visible wavelengths are intensified.
P37.29
(a)
For maximum transmission, we want destructive interference in the light reflected from the
front and back surfaces of the film.
If the surrounding glass has refractive index greater than 1.378, light reflected from the
front surface suffers no phase reversal and light reflected from the back does undergo phase
reversal. This effect by itself would produce destructive interference, so we want the
λ
distance down and back to be one whole wavelength in the film: 2t = .
n
λ 656.3 nm
t=
=
= 238 nm
2n 2 (1.378 )
(b)
The filter will undergo thermal expansion. As t increases in 2nt = λ , so does λ increase .
(c)
Destructive interference for reflected light happens also for λ in 2nt = 2λ ,
or
λ = 1.378 ( 238 nm ) = 328 nm
( near ultraviolet )
Interference of Light Waves
P37.30
If the path length difference ∆ = λ , the transmitted light will be bright. Since ∆ = 2d = λ ,
dmin =
P37.31
363
λ 580 nm
=
= 290 nm
2
2
For destructive interference in the air,
2t = mλ
For 30 dark fringes, including the one where the plates
meet,
29 ( 600 nm )
= 8.70 × 10 −6 m
2
Therefore, the radius of the wire is
t=
r=
t 8.70 µ m
=
= 4.35 µ m
2
2
FIG. P37.31
P37.32
The condition for bright fringes is
2t +
λ
λ
=m
2n
n
m = 1, 2, 3, . . .
θ
R
From the sketch, observe that
⎛
θ2 ⎞ R r
r2
t = R (1 − cos θ ) ≈ R ⎜ 1 − 1 + ⎟ = ⎛ ⎞ =
⎝
2 ⎠ 2 ⎝ R⎠
2R
2
P37.33
t
The condition for a bright fringe becomes
r2 ⎛
1 λ
= ⎜ m − ⎞⎟
R ⎝
2⎠ n
Thus, for fixed m and λ ,
nr 2 = constant
Therefore, nliquid rf2 = nair ri2 and
nliquid = (1.00 )
r
FIG. P37.32
(1.50 cm )2
= 1.31
(1.31 cm )2
For total darkness, we want destructive interference for reflected light for both 400 nm and 600 nm.
With phase reversal at just one reflecting surface, the condition for destructive interference is
2nair t = mλ
m = 0, 1, 2, . . .
The least common multiple of these two wavelengths is 1 200 nm, so we get no reflected light at
2 (1.00 ) t = 3 ( 400 nm ) = 2 ( 600 nm ) = 1 200 nm, so t = 600 nm at this second dark fringe.
By similar triangles,
or the distance from the contact point is
600 nm 0.050 0 mm
=
x
10.0 cm
0.100 m ⎞
x = ( 600 × 10 −9 m ) ⎛
= 1.20 mm
⎝ 5.00 × 10 −5 m ⎠
364
Chapter 37
Section 37.7
P37.34
The Michelson Interferometer
Distance = 2 ( 3.82 × 10 −4 m ) = 1 700 λ
λ = 4.49 × 10 −7 m = 449 nm
The light is blue .
P37.35
When the mirror on one arm is displaced by ∆, the path difference changes by 2∆. A shift
resulting in the reversal between dark and bright fringes requires a path length change of one-half
mλ
wavelength. Therefore, 2 ∆ =
, where in this case, m = 250.
2
−7
λ ( 250 ) ( 6.328 × 10 m )
∆ = m =
= 39.6 µ m
4
4
P37.36
Counting light going both directions, the number of wavelengths originally in the cylinder is
2L
2 L 2nL
=
m1 =
. It changes to m2 =
as the cylinder is filled with gas. If N is the number of
λ
λ n
λ
2L
bright fringes passing, N = m2 − m1 =
( n − 1), or the index of refraction of the gas is
λ
Nλ
n = 1+
2L
Additional Problems
*P37.37 The same light source will radiate into the syrup light with wavelength 560 nm/1.38 = 406 nm.
The first side bright fringe is separated from the central bright fringe by distance y described by
d sin θ = 1λ
P37.38
dy/ L ≈ λ
y = λ L / d = 406 × 10 −9 m(1.20 m)/(30 × 100 −6 m) = 1.62 cm
(a)
Where fringes of the two colors coincide we have d sin θ = mλ = m ′λ ′, requiring
(b)
λ = 430 nm, λ ′ = 510 nm
∴
λ m′
.
=
λ′ m
m ′ 430 nm 43
=
= , which cannot be reduced any further. Then m = 51 , m ′ = 43.
m 510 nm 51
⎡ ( 51) ( 430 × 10 −9 m ) ⎤
⎛ mλ ⎞
−1
θ m = sin −1 ⎜
sin
=
⎢
⎥ = 61.3°
−3
⎝ d ⎟⎠
⎢⎣ 0.025 × 10 m ⎥⎦
ym = L tan θ m = (1.5 m ) tan 61.3° = 2.74 m
P37.39
c 3.00 × 108 m s
= 5.00 m.
=
f
60.0 × 10 6 s −1
Along the line AB the two traveling waves going in opposite directions add to give a standing
wave. The two transmitters are exactly 2.00 wavelengths apart and the signal from B, when it
arrives at A, will always be in phase with transmitter B. Since B is 180° out of phase with A, the
two signals always interfere destructively at the position of A.
The wavelength is λ =
The first antinode (point of constructive interference) is located at distance
λ 5.00 m
=
= 1.25 m from the node at A.
4
4
Interference of Light Waves
P37.40
Along the line of length d joining the source, two identical waves
moving in opposite directions add to give a standing wave.
s
N
A
365
N
λ
d λ
4
An antinode is halfway between the sources. If > , there is
2 2
FIG. P37.40
d
space for two more antinodes for a total of three. If > λ , there
2
d
will be at least five antinodes, and so on. To repeat, if > 0, the
λ
d
number of antinodes is 1 or more. If > 1, the number of
λ
d
antinodes is 3 or more. If > 2, the number of antinodes is 5 or more. In general,
λ
the number of antinodes is 1 plus 2 times thhe greatest integer less than or equal to
d
.
λ
d λ
d λ
< , there will be no nodes. If > , there will be space for at least two nodes, as shown
2 4
2 4
d 3λ
d 5λ
in the picture. If >
, there will be at least four nodes. If >
, six or more nodes will fit
2
4
2
4
in, and so on. To repeat, if 2d < λ , the number of nodes is 0. If 2d > λ , the number of nodes is 2
If
or more. If 2d > 3λ , the number of nodes is 4 or more. If 2d > 5λ , the number of nodes is 6 or
d 1
d 1
more. Again, if ⎛ + ⎞ > 1, the number of nodes is at least 2. If ⎛ + ⎞ > 2, the number of
⎝ λ 2⎠
⎝ λ 2⎠
d 1
nodes is at least 4. If ⎛ + ⎞ > 3, the number of nodes is at least 6. In general,
⎝ λ 2⎠
d 1
the number of nodes is 2 times the greatest nonzero integer less than ⎛ + ⎞ .
⎝ λ 2⎠
Next, we enumerate the zones of constructive interference. They are described by d sin θ = mλ ,
m = 0, 1, 2, . . . with θ counted as positive both left and right of the maximum at θ = 0 in the
center. The number of side maxima on each side is the greatest integer satisfying sin θ ≤ 1,
d
d1 ≥ mλ , m ≤ . So the total
λ
number of bright fringes is one plus 2 timess the greatest integer less than or equal to
d
.
λ
It is equal to the number of antinodes on the line joining the sources.
1
The interference minima are to the left and right at angles described by d sin θ = ⎛ m + ⎞ λ ,
⎝
2⎠
1⎞
d 1
d 1
⎛
m = 0, 1, 2, . . . . With sin θ < 1, d1 > mmax +
λ , mmax < − or mmax + 1 < + . Let
⎝
λ 2
2⎠
λ 2
d 1
n = 1, 2, 3, . . . . Then the number of side minima is the greatest integer n less than + .
λ 2
Counting both left and right,
d 1
the number of dark fringes is two times the greatest positive integer less than ⎛ + ⎞ .
⎝ λ 2⎠
It is equal to the number of nodes in the standing wave between the sources.
s
366
P37.41
Chapter 37
My middle finger has width d = 2 cm.
(a)
Two adjacent directions of constructive interference for 600-nm light are described by
d sin θ = mλ
θ0 = 0
( 2 × 10
(b)
−2
m ) sin θ1 = 1 ( 6 × 10 −7 m )
Thus,
θ1 = 2 × 10 −3 degree
and
θ1 − θ 0 ~10 −3 degree
Choose
θ1 = 20°
( 2 × 10
−2
m ) sin 20° = (1) λ
λ = 7 mm
Millimeter waves are microwaves .
f =
c
:
λ
f =
3 × 108 m s
~1011 Hz
7 × 10 −3 m
*P37.42 Constructive interference occurs where m = 0, 1, 2, 3, . . . , −1, −2, −3, . . . , for
⎛ 2π x1 − 2π ft + π ⎞ − ⎛ 2π x2 − 2π ft + π ⎞ = 2π m
⎝ λ
6⎠ ⎝ λ
8⎠
( x1 − x2 ) +
650 nm
2π ( x1 − x2 ) ⎛ π π ⎞
+
−
= 2π m
⎝ 6 8⎠
λ
1
1
1
−
= m x1 − x2 = ⎛ m − ⎞ 650 nm with m = 0, 1, 2, 3, . . . , − 1, − 2, − 3, . . .
⎝
12 16
48 ⎠
*P37.43 A bright line for the green light requires dy/L = m1λ 1. A blue interference maximum requires
dy/L = m2λ 2 for integers m1 and m2. Then m1 540 nm = m2 450 nm. The smallest integers
satisfying the equation are m1 = 5 and m2 = 6. Then for both
dy/L = 2 700 nm
P37.44
y = (1.4 m) 2.7 µ m/ 150 µ m = 2.52 cm
If the center point on the screen is to be a dark spot rather than bright, passage through the plastic
must delay the light by one-half wavelength. Calling the thickness of the plastic t,
t 1
t
nt
+ =
=
λ 2 λ n λ
or
t=
λ
where n is the index of refraction for the plastic.
2 ( n − 1)
Interference of Light Waves
P37.45
No phase shift upon reflection from the upper surface (glass to air) of the film, but there will be a
λ
shift of due to the reflection at the lower surface of the film (air to metal). The total phase
2
difference in the two reflected beams is
δ = 2nt +
then
λ
2
For constructive interference, δ = mλ
2 (1.00 ) t +
or
λ
= mλ
2
Thus, the film thickness for the mth order bright fringe is
1 λ
⎛ λ⎞ λ
tm = ⎛ m − ⎞ = m ⎜ ⎟ −
⎝
⎝ 2⎠ 4
2⎠ 2
and the thickness for the m − 1 bright fringe is:
⎛ λ⎞ λ
t m−1 = ( m − 1) ⎜ ⎟ −
⎝ 2⎠ 4
Therefore, the change in thickness required to go from one bright fringe to the next is
∆t = t m − t m −1 =
λ
2
To go through 200 bright fringes, the change in thickness of the air film must be:
⎛ λ⎞
200 ⎜ ⎟ = 100 λ
⎝ 2⎠
Thus, the increase in the length of the rod is
∆L = 100 λ = 100 ( 5.00 × 10 −7 m ) = 5.00 × 10 −5 m
P37.46
367
From
∆L = Liα ∆T
we have:
α=
∆L
5.00 × 10 −5 m
=
= 20.0 × 10 −6 °C−1
Li ∆T ( 0.100 m ) ( 25.0 ° C )
Since 1 < 1.25 < 1.34, light reflected from top and bottom surfaces of the oil undergoes phase
reversal. The path difference is then 2t, which must be equal to
mλ n =
mλ
n
for maximum reflection, with m = 1 for the given first-order condition and n = 1.25. So
t=
mλ 1 ( 500 nm )
=
= 200 nm
2n
2 (1.25)
The volume we assume to be constant:
A=
1.00 m 3 = ( 200 nm ) A
1.00 m 3
= 5.00 × 10 6 m 2 = 5.00 km 2
200 (10 −9 m )
368
P37.47
Chapter 37
One radio wave reaches the receiver R directly from the
distant source at an angle θ above the horizontal. The other
wave undergoes phase reversal as it reflects from the water
at P.
Constructive interference first occurs for a path difference of
λ
(1)
d=
2
It is equally far from P to R as from P to R ′, the mirror
image of the telescope.
The angles θ in the figure are equal because they each form
part of a right triangle with a shared angle at R ′.
d = 2 ( 20.0 m ) sin θ = ( 40.0 m ) sin θ
So the path difference is
Substituting for d and λ in Equation (1),
P37.48
5.00 m
and θ = 3.58° .
80.0 m
For destructive interference, the path length must differ by mλ . We may treat this problem as
π
a double slit experiment if we remember the light undergoes a -phase shift at the mirror. The
2
second slit is the mirror image of the source, 1.00 cm below the mirror plane. Modifying
Equation 37.7,
ydark =
P37.49
c 3.00 × 108 m s
=
= 5.00 m
f
60.0 × 10 6 Hz
5.00 m
( 40.0 m ) sin θ =
2
λ=
The wavelength is
Solving for the angle θ , sin θ =
FIG. P37.47
−7
mλ L 1 ( 5.00 × 10 m ) (100 m )
=
= 2.50 mm
d
( 2.00 × 10 −2 m )
2 (15.0 km ) + h 2 = 30.175 km
2
(15.0 km )2 + h 2 = 227.63
h = 1.62 km
P37.50
For dark fringes,
and at the edge of the wedge,
When submerged in water,
m=
so
FIG. P37.49
2nt = mλ
84 ( 500 nm )
2
2nt = mλ
t=
2 (1.33) ( 42 ) ( 500 nm )
500 nm
m + 1 = 113 dark fringes
FIG. P37.50
Interference of Light Waves
P37.51
⎛ π yd ⎞
I
= cos 2 ⎜
⎝ λ L ⎟⎠
I max
From Equation 37.14,
I
Let λ2 equal the wavelength for which
I max
λ2 =
Then
But
⎛ I ⎞
π yd
= λ1 cos −1 ⎜ 1 ⎟
L
⎝ I max ⎠
→
I2
I max
= 0.640
π yd L
12
cos ( I 2 I max )
−1
12
= ( 600 nm ) cos −1 ( 0.900 ) = 271 nm
Substituting this value into the expression for λ2,
⎛ I ⎞
Note that in this problem, cos −1 ⎜
⎝ I max ⎟⎠
P37.52
369
λ2 =
12
At entrance, 1.00 sin 30.0° = 1.38 sin θ 2
Call t the unknown thickness. Then
271 nm
= 421 nm
cos −1 ( 0.6401 2 )
must be expressed in radians.
θ 2 = 21.2°
t
t
a=
a
cos 21.2°
c
tan 21.2° =
c = t tan 21.2°
t
b
FIG. P37.52
sin θ1 =
b = 2t tan 21.2° sin 30.0°
2c
The net shift for the second ray, including the phase reversal on reflection of the first, is
cos 21.2° =
2an − b −
λ
2
where the factor n accounts for the shorter wavelength in the film. For constructive interference,
we require
The minimum thickness will be given by
2an − b −
λ
= mλ
2
2an − b −
λ
=0
2
λ
nt
= 2an − b = 2
− 2t (tan 21.2°) sin 30.0°
2
cos 21.2°
590 nm ⎛ 2 × 1.38
=
− 2 tan 21.2° sin 30.0° ⎞ t = 2.57t
⎝ cos 21.2°
⎠
2
t = 115 nm
370
P37.53
Chapter 37
The shift between the two reflected waves is δ = 2na − b −
λ
2
where a and b are as shown in the ray diagram, n is the
λ
index of refraction, and the term
is due to phase reversal
2
at the top surface. For constructive interference, δ = mλ
where m has integer values. This condition becomes
1
2na − b = ⎛ m + ⎞ λ
⎝
2⎠
From the figure’s geometry,
a=
FIG. P37.53
(1)
t
cos θ 2
c = a sin θ 2 =
t sin θ 2
cos θ 2
b = 2c sin φ1 =
Also, from Snell’s law,
sin φ1 = n sin θ 2
Thus,
b=
2t sin θ 2
sin φ1
cos θ 2
2nt sin 2 θ 2
cos θ 2
With these results, the condition for constructive interference given in Equation (1) becomes:
⎛ t ⎞ 2nt sin 2 θ 2
2nt
1
2n ⎜
1 − sin 2 θ 2 ) = ⎛ m + ⎞ λ
−
=
(
⎟
⎝
cos θ 2
cos θ 2
2⎠
⎝ cos θ 2 ⎠
1
2nt cos θ 2 = ⎛ m + ⎞ λ
⎝
2⎠
or
*P37.54 (a)
For a linear function taking the value 1.90 at y = 0 and 1.33 at y = 20 cm we write
n( y) = 1.90 + (1.33 − 1.90)y/20 cm
(b)
∫
20 cm
0
n( y)dy =
∫
20 cm
0
or
n ( y) = 1.90 − 0.0285 y/cm
0.0285 y 2
[1.90 − 0.0285 y / cm]dy = 1.90 y −
2
= 38.0 cm − 5.7 cm
= 32.3 cm
(c)
The beam will continuously curve downward.
20 cm
0
Interference of Light Waves
P37.55
(a)
Minimum: 2nt = mλ2
for m = 0, 1, 2, . . .
1
Maximum: 2nt = ⎛ m ′ + ⎞ λ1
⎝
2⎠
for m ′ = 0, 1, 2, . . .
for λ1 > λ2,
⎛ m′ + 1⎞ < m
⎝
2⎠
so
m′ = m − 1
Then
1
2nt = mλ2 = ⎛ m − ⎞ λ1
⎝
2⎠
371
2 mλ2 = 2 mλ1 − λ1
m=
so
(b)
m=
λ1
2 ( λ1 − λ2 )
500
= 1.92 → 2 (wavelengths measured to ±5 nm)
2 ( 500 − 370 )
Minimum:
2nt = mλ2
2 (1.40 ) t = 2 ( 370 nm )
Maximum:
t = 264 nm
1
2nt = ⎛ m − 1 + ⎞ λ = 1.5λ
⎝
2⎠
2 (1.40 ) t = 1.5 ( 500 nm )
t = 268 nm
Film thickness = 266 nm
P37.56
From the sketch, observe that
4h2 + d 2
2
Including the phase reversal due to reflection from the ground, the total shift between
λ
the two waves is δ = 2 x − d − .
2
d
2
d
x = h2 + ⎛ ⎞ =
⎝ 2⎠
x
h
x
d/2
FIG. P37.56
(a)
(b)
For constructive interference, the total shift must be an integral number of wavelengths, or
δ = mλ where m = 0, 1, 2, 3, . . .
1
4 x − 2d
Thus,
2 x − d = ⎛ m + ⎞ λ or
λ=
⎝
⎠
2
2m + 1
For the longest wavelength, m = 0, giving
λ = 4 x − 2d = 2 4 h 2 + d 2 − 2d
For destructive interference,
1
δ = ⎛ m − ⎞ λ where m = 0, 1, 2, 3, . . . .
⎝
2⎠
Thus,
2x − d = mλ
For the longest wavelength, m = 1 giving
λ = 2x − d =
or
4h2 + d 2 − d
λ=
2x − d
m
372
P37.57
Chapter 37
Call t the thickness of the film. The central maximum
corresponds to zero phase difference. Thus, the added
distance ∆r traveled by the light from the lower slit
must introduce a phase difference equal to that introduced by the plastic film. The phase difference φ is
L
Thin
film
y'
d
⎛ t ⎞
φ = 2π ⎜ ⎟ ( n − 1)
⎝ λa ⎠
The corresponding difference in path length ∆r is
θ
α
Zero
order
m=0
Screen
∆r
FIG. P37.57
⎛ t ⎞
⎛λ ⎞
⎛λ ⎞
∆r = φ ⎜ a ⎟ = 2π ⎜ ⎟ ( n − 1) ⎜ a ⎟ = t ( n − 1)
⎝ 2π ⎠
⎝ 2π ⎠
⎝ λa ⎠
Note that the wavelength of the light does not appear in this equation. In the figure, the two rays
from the slits are essentially parallel.
∆r y ′
Thus the angle θ may be expressed as
tan θ =
=
d
L
t ( n − 1) L
y ′ t ( n − 1)
Eliminating ∆r by substitution,
gives y ′ =
=
d
L
d
P37.58
2t =
and dark fringes occur when
⎛ λ⎞
2t = ⎜ ⎟ m
⎝ n⎠
The thickness of the film at x is
h
t=⎛ ⎞x
⎝ ⎠
Therefore, xbright =
P37.59
λ⎛
1
m+ ⎞
n⎝
2⎠
Bright fringes occur when
(a)
λ m
λ ⎛
1
.
m + ⎞ and xdark =
⎝
⎠
2hn
2hn
2
FIG. P37.58
1⎞
⎛
λ . The first bright ring
Constructive interference in the reflected light requires 2t = m +
⎝
2⎠
has m = 0 and the 55th has m = 54, so at the edge of the lens
t=
54.5 ( 650 × 10 −9 m )
2
= 17.7 µ m
Now from the geometry in textbook Figure 37.12, we can find the distance t from the
curved surface down to the flat plate by considering distances down from the center
of curvature:
R 2 − r 2 = R − t or R 2 − r 2 = R 2 − 2 Rt + t 2
−2
−5
r 2 + t 2 ( 5.00 × 10 m ) + (1.77 × 10 m )
=
R=
= 70.6 m
−5
2t
2 (1.77 × 10 m )
2
(b)
2
⎛ 1
1
1⎞
1
1
⎞ so f = 136 m
− ⎟ = 0.520 ⎛ −
= ( n − 1) ⎜
⎝ ∞ −70.6 m ⎠
f
⎝ R2 R2 ⎠
Interference of Light Waves
P37.60
The shift between the waves reflecting from the top and
bottom surfaces of the film at the point where the film has
λ
thickness t is δ = 2tnfilm + , with the factor of λ being due
2
2
to a phase reversal at one of the surfaces.
For the dark rings (destructive interference), the total shift
θ
P37.61
t
r
mλ
This requires that t =
.
2nfilm
FIG. P37.60
To find t in terms of r and R,
R2 = r 2 + ( R − t )
Since t is much smaller than R,
t 2 << 2 Rt
Thus, where m is an integer,
r≈
2
so
r 2 = 2 Rt + t 2
and
⎛ mλ ⎞
r 2 ≈ 2 Rt = 2 R ⎜
⎝ 2nfilm ⎟⎠
mλ R
nfilm
Light reflecting from the upper interface of the air layer suffers
no phase change, while light reflecting from the lower interface
is reversed 180°. Then there is indeed a dark fringe at the outer
circumference of the lens, and a dark fringe wherever the air
thickness t satisfies 2t = mλ , m = 0, 1, 2, . . . .
(a)
R
nfilm
1
⎛
should be δ = m + ⎞ λ with m = 0, 1, 2, 3, . . . .
⎝
2⎠
t0
r
At the central dark spot m = 50 and
R=8m
50 λ
= 25 ( 589 × 10 −9 m ) = 1.47 × 10 −5 m
2
In the right triangle,
t0 =
(b)
(8 m )2 = r 2 + (8 m − 1.47 × 10 −5 m ) = r 2 + (8 m )2
2
− 2 (8 m ) (1.47 × 10 −5 m ) + 2 × 10 −10 m 2
FIG. P37.61
The last term is negligible. r = 2 (8 m ) (1.47 × 10 −5 m ) = 1.53 × 10 −2 m
(c)
⎛ 1
1
1⎞
1
1 ⎞
= ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎛ −
⎝
f
∞ 8.00 m ⎠
⎝ R1 R2 ⎠
f = −16.0 m
373
374
P37.62
Chapter 37
For bright rings the gap t between surfaces is given by
1
2t = ⎛ m + ⎞ λ . The first bright ring has m = 0 and the
⎝
2⎠
hundredth has m = 99.
1
So, t = ( 99.5) ( 500 × 10 −9 m ) = 24.9 µ m.
2
Call rb the ring radius. From the geometry of the figure
at the right,
(
t = r − r 2 − rb2 − R − R 2 − rb2
)
Since rb << r , we can expand in series:
FIG. P37.62
⎛
⎛
1 r2 ⎞
1 r2 ⎞ 1 r2 1 r2
t = r − r ⎜ 1 − b2 ⎟ − R + R ⎜ 1 − b2 ⎟ = b − b
⎝
⎝
2r ⎠
2R ⎠ 2 r 2 R
12
⎡ 2t
⎤
rb = ⎢
⎥
⎣1 r − 1 R ⎦
P37.63
(a)
⎡ 2 ( 24.9 × 10 −6 m ) ⎤
=⎢
⎥
⎢⎣ 1 4.00 m − 1 12.0 m ⎥⎦
12
= 1.73 cm
1
⎛
Bright bands are observed when 2nt = m + ⎞ λ .
⎝
2⎠
Hence, the first bright band (m = 0) corresponds to nt =
(b)
(c)
P37.64
Since
x1 t1
=
x2 t 2
we have
⎛λ ⎞
⎛t ⎞
m⎞
680 nm
x2 = x1 ⎜ 2 ⎟ = x1 ⎜ 2 ⎟ = ( 3.00 cm ) ⎛
= 4.86 cm
⎝
420 nm ⎠
⎝ λ1 ⎠
⎝ t1 ⎠
t1 =
λ1 420 nm
= 78.9 nm
=
4 n 4 (1.33)
t2 =
λ2 680 nm
=
= 128 nm
4 n 4 (1.33)
θ ≈ tan θ =
t1 78.9 nm
=
= 2.63 × 10 −6 rad
x1 3.00 cm
Depth = one-quarter of the wavelength in plastic.
t=
P37.65
λ
.
4
λ 780 nm
=
= 130 nm
4 n 4 (1.50 )
1
2h sin θ = ⎛ m + ⎞ λ
⎝
2⎠
∆y
1
2h ⎛ ⎞ = λ
⎝ 2L ⎠ 2
bright
so
h=
−9
L λ ( 2.00 m ) ( 606 × 10 m )
=
= 0.505 mm
2 ∆y
2 (1.2 × 10 −3 m )
Interference of Light Waves
P37.66
375
Superposing the two vectors, ER = E1 + E2
2
2
E
E2
E2
2
⎛
⎛E
ER = E1 + E 2 = E0 + 0 cos φ ⎞ + 0 sin φ ⎞ = E02 + E02 cos φ + 0 cos 2 φ + 0 sin 2 φ
⎝
⎠ ⎝ 3
⎠
3
9
9
3
ER =
10 2 2 2
E0 + E0 cos φ
9
3
Since intensity is proportional to the square of the amplitude,
I=
10
2
I max + I max cos φ
9
3
Using the trigonometric identity cos φ = 2 cos 2
I=
or
P37.67
φ
− 1, this becomes
2
φ
φ
10
2
4
4
I max + I max ⎛ 2 cos 2 − 1⎞ = I max + I max cos 2
⎝
⎠
2
9
3
2
9
3
I=
φ
4
I max ⎛ 1 + 3 cos 2 ⎞
⎝
9
2⎠
Represent the light radiated from each slit to point P as a phasor. The two
have equal amplitudes E. Since intensity is proportional to amplitude
squared, they add to amplitude 3E.
3E 2
, θ = 30° . Next, the obtuse angle
E
between the two phasors is 180 − 30 − 30 = 120°, and
Then cos θ =
φ = 180 − 120°= 60° . The phase difference between the
two phasors is caused by the path difference δ = SS2 − SS1
according to
δ
φ
60°
λ
, δ=λ
=
= . Then
λ 360°
360°
6
λ
6
2
Lλ λ 2
L2 + d 2 = L2 +
+
6
36
L2 + d 2 − L =
The last term is negligible, so
⎛ 2 Lλ ⎞
d=⎜
⎝ 6 ⎟⎠
12
=
2 (1.2 m ) 620 × 10 −9 m
= 0.498 mm
6
3E
θ
φ
E
FIG. P37.67
E
376
Chapter 37
ANSWERS TO EVEN PROBLEMS
P37.2
515 nm
P37.4
(a) 36.2°
P37.6
Question: A single oscillator makes the two speakers of a boom box, 35.0 cm apart, vibrate in
phase at 1.62 kHz. At what angles, measured from the perpendicular bisector of the line joining
the speakers, will a distant observer hear maximum sound intensity? Minimum sound intensity?
The ambient temperature is 20°C. Answer: The wavelength of the sound is 21.2 cm. Interference
maxima occur at angles of 0° and 37.2° to the left and right. Minima occur at angles of 17.6° and
65.1°. No second-order or higher-order maximum exists. No angle exists, smaller or larger than
90°, for which sinq2 loud = 1.21. No location exists in the Universe that is two wavelengths farther
from one speaker than from the other.
P37.8
11.3 m
P37.10
641
P37.12
6.33 mm/s
P37.14
See the solution.
P37.16
(a) 1.29 rad
P37.18
0.968
P37.20
See the solution.
P37.22
(a) See the solution. (b) The cosine function takes on the extreme value −1 to describe the
secondary maxima. The cosine function takes on the extreme value +1 to describe the primary
maxima. The ratio is 9.00.
P37.24
612 nm
P37.26
0.500 cm
P37.28
No reflection maxima in the visible spectrum
P37.30
290 nm
P37.32
1.31
P37.34
449 nm; blue
P37.36
1 + Nl/2L
P37.38
(a) See the solution.
(b) 5.08 cm
(c) 508 THz
(b) 99.6 nm
(b) 2.74 m
Interference of Light Waves
P37.40
377
number of antinodes = number of constructive interference zones
= 1 plus 2 times the greatest positive integer ≤
d
λ
number of nodes = number of destructive interference zones
d 1
= 2 times the greatest positive integer < ⎛ + ⎞
⎝ λ 2⎠
P37.42
1
x1 − x2 = ⎛ m − ⎞ 650 nm with m = 0, 1, 2, 3, . . . , − 1, − 2, − 3, . . .
⎝
48 ⎠
P37.44
λ
2 ( n − 1)
P37.46
5.00 km 2
P37.48
2.50 mm
P37.50
113
P37.52
115 nm
P37.54
(a) n(y) = 1.90 − 0.0285 y/cm (b) 32.3 cm
P37.56
(a) 2(4h2 + d 2)1/2 − 2d
P37.58
See the solution.
P37.60
See the solution.
P37.62
1.73 cm
P37.64
130 nm
P37.66
See the solution.
(b) (4h2 + d 2)1/2 − d
(c) The beam will continuously curve downward.
38
Diffraction Patterns and Polarization
CHAPTER OUTLINE
38.1
38.2
38.3
38.4
38.5
38.6
Introduction to Diffraction
Patterns
Diffraction Patterns from Narrow
Slits
Resolution of Single-Slit and
Circular Apertures
The Diffraction Grating
Diffraction of X-Rays by Crystals
Polarization of Light Waves
Q38.2
ANSWERS TO QUESTIONS
Q38.1
Audible sound has wavelengths on the order of meters
or centimeters, while visible light has a wavelength
on the order of half a micrometer. In this world of
λ
breadbox-sized objects, is large for sound, and sound
a
λ
diffracts around behind walls with doorways. But
a
is a tiny fraction for visible light passing ordinary-size
objects or apertures, so light changes its direction by
only very small angles when it diffracts.
Another way of phrasing the answer: We can see by a
small angle around a small obstacle or around the edge
of a small opening. The side fringes in Figure 38.1 and
the Arago spot in the center of Figure 38.3 show this
diffraction. We cannot always hear around corners.
Out-of-doors, away from reflecting surfaces, have someone a few meters distant face away from you and whisper.
The high-frequency, short-wavelength, informationcarrying components of the sound do not diffract around
his head enough for you to understand his words.
Suppose an opera singer loses the tempo and cannot
immediately get it from the orchestra conductor. Then
the prompter may make rhythmic kissing noises with her
lips and teeth. Try it—you will sound like a birdwatcher
trying to lure out a curious bird. This sound is clear on
the stage but does not diffract around the prompter’s box
enough for the audience to hear it.
The wavelength of visible light is extremely small in comparison to the dimensions of your hand,
so the diffraction of light around an obstacle the size of your hand is totally negligible. However,
sound waves have wavelengths that are comparable to the dimensions of the hand or even larger.
Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.
*Q38.3 Answer (d). The power of the light coming through the slit decreases, as you would expect. The
λ
central maximum increases in width as the width of the slit decreases. In the condition sin θ =
a
for destructive interference on each side of the central maximum, θ increases as a decreases.
*Q38.4 We consider the fraction λ L/a. In case (a) and in case (f) it is λ 0L/a. In case (b) it is 2λ 0L/3a.
In case (c) it is 3λ 0L/2a. In case (d) it is l0L/2a. In case (e) it is l02L/a. The ranking is e > c > a =
f > b > d.
*Q38.5 Answer (b). The wavelength will be much smaller than with visible light.
379
380
Chapter 38
*Q38.6 Answer (c). The ability to resolve light sources depends on diffraction, not on intensity.
Q38.7
Consider incident light nearly parallel to the horizontal
ruler. Suppose it scatters from bumps at distance d apart
to produce a diffraction pattern on a vertical wall a
y
distance L away. At a point of height y, where θ =
L
gives the scattering angle θ , the character of the
interference is determined by the shift δ between beams
d
− d.
scattered by adjacent bumps, where δ =
cos θ
y
q
d
L
FIG. Q38.7
Bright spots appear for δ = mλ , where m = 0, 1, 2, 3, . . . . For small θ , these equations combine
y2 d
and reduce to mλ = m 2 . Measurement of the heights ym of bright spots allows calculation of the
2L
wavelength of the light.
*Q38.8 Answer (b). No diffraction effects are observed because the separation distance between adjacent
ribs is so much greater than the wavelength of x-rays. Diffraction does not limit the resolution of
an x-ray image. Diffraction might sometimes limit the resolution of an ultrasonogram.
*Q38.9 Answer (a). Glare, as usually encountered when driving or boating, is horizontally polarized.
Reflected light is polarized in the same plane as the reflecting surface. As unpolarized light hits
a shiny horizontal surface, the atoms on the surface absorb and then reemit the light energy as
a reflection. We can model the surface as containing conduction electrons free to vibrate easily along the surface, but not to move easily out of surface. The light emitted from a vibrating
electron is partially or completely polarized along the plane of vibration, thus horizontally.
Q38.10 Light from the sky is partially polarized. Light from the blue sky that is polarized at 90° to the
polarization axis of the glasses will be blocked, making the sky look darker as compared to the
clouds.
*Q38.11 Answer (a). The grooves in a diffraction grating are not electrically conducting. Sending light
through a diffraction grating is not like sending a vibration on a rope through a picket fence. The
electric field in light does not have an amplitude in real space. Its amplitude is in newtons per
coulomb, not in millimeters.
Q38.12 First think about the glass without a coin and about one particular point P on the screen. We can
divide up the area of the glass into ring-shaped zones centered on the line joining P and the light
source, with successive zones contributing alternately in-phase and out-of-phase with the light
that takes the straight-line path to P. These Fresnel zones have nearly equal areas. An outer zone
contributes only slightly less to the total wave disturbance at P than does the central circular
zone. Now insert the coin. If P is in line with its center, the coin will block off the light from
some particular number of zones. The first unblocked zone around its circumference will send
light to P with significant amplitude. Zones farther out will predominantly interfere destructively
with each other, and the Arago spot is bright. Slightly off the axis there is nearly complete
destructive interference, so most of the geometrical shadow is dark. A bug on the screen crawling
out past the edge of the geometrical shadow would in effect see the central few zones coming out
of eclipse. As the light from them interferes alternately constructively and destructively, the bug
moves through bright and dark fringes on the screen. The diffraction pattern is shown in
Figure 38.3 in the text.
Diffraction Patterns and Polarization
381
Q38.13 Since the obsidian is opaque, a standard method of measuring incidence and refraction angles
and using Snell’s Law is ineffective. Reflect unpolarized light from the horizontal surface
of the obsidian through a vertically polarized filter. Change the angle of incidence until you
observe that none of the reflected light is transmitted through the filter. This means that the
reflected light is completely horizontally polarized, and that the incidence and reflection angles
are the polarization angle. The tangent of the polarization angle is the index of refraction of the
obsidian.
Q38.14 The fine hair blocks off light that would otherwise go through a fine slit and produce a diffraction
pattern on a distant screen. The width of the central maximum in the pattern is inversely
proportional to the distance across the slit. When the hair is in place, it subtracts the same
diffraction pattern from the projected disk of laser light. The hair produces a diffraction minimum
that crosses the bright circle on the screen. The width of the minimum is inversely proportional
to the diameter of the hair. The central minimum is flanked by narrower maxima and minima.
Measure the width 2y of the central minimum between the maxima bracketing it, and use
y λ
Equation 38.1 in the form = to find the width a of the hair.
L a
Q38.15 The condition for constructive interference is that the three radio signals arrive at the city in
phase. We know the speed of the waves (it is the speed of light c), the angular bearing θ of the
city east of north from the broadcast site, and the distance d between adjacent towers. The wave
from the westernmost tower must travel an extra distance 2d sin θ to reach the city, compared to
the signal from the eastern tower. For each cycle of the carrier wave, the western antenna would
d sin θ
transmit first, the center antenna after a time delay
, and the eastern antenna after an
c
additional equal time delay.
Q38.16 It is shown in the correct orientation. If the horizontal width of the opening is equal to or less than
the wavelength of the sound, then the equation a sin θ = (1) λ has the solution θ = 90°, or has no
solution. The central diffraction maximum covers the whole seaward side. If the vertical height
of the opening is large compared to the wavelength, then the angle in a sin θ = (1) λ will be small,
and the central diffraction maximum will form a thin horizontal sheet.
Featured in the motion picture M*A*S*H (20th Century Fox, Aspen Productions, 1970)
is a loudspeaker mounted on an exterior wall of an Army barracks. It has an approximately
rectangular aperture, and it is installed incorrectly. The longer side is horizontal, to maximize
sound spreading in a vertical plane and to minimize sound radiated in different horizontal
directions.
SOLUTIONS TO PROBLEMS
Section 38.1
Introduction to Diffraction Patterns
Section 38.2
Diffraction Patterns from Narrow Slits
P38.1
sin θ =
λ 6.328 × 10 −7
= 2.11 × 10 −3
=
a
3.00 × 10 −4
y
= tan θ ≈ sin θ ≈ θ (for small θ )
1.00 m
2 y = 4.22 mm
382
P38.2
Chapter 38
The positions of the first-order minima are
y
λ
≈ sin θ = ± . Thus, the spacing between these two
L
a
⎛ λ⎞
minima is ∆y = 2 ⎜ ⎟ L and the wavelength is
⎝ a⎠
⎛ 4.10 × 10 −3 m ⎞ ⎛ 0.550 × 10 −3 m ⎞
∆y a
λ =⎛ ⎞⎛ ⎞ =⎜
⎟⎠ ⎜⎝
⎟⎠ = 547 nm
⎝ 2 ⎠ ⎝ L⎠ ⎝
2.06 m
2
P38.3
y
mλ
≈ sin θ =
L
a
∆m = 3 − 1 = 2
∆y = 3.00 × 10 −3 nm
and
a=
a=
P38.4
∆mλ L
∆y
2 ( 690 × 10 −9 m ) ( 0.500 m )
(3.00 × 10 −3 m )
= 2.30 × 10 −4 m
For destructive interference,
λ λ 5.00 cm
sin θ = m = =
= 0.139
a a 36.0 cm
θ = 7.98°
and
y
= tan θ
L
gives
y = L tan θ = ( 6.50 m ) tan 7.98° = 0.912 m
y = 91.2 cm
P38.5
v 340 m s
=
= 0.523 m
f
650 s −1
If the speed of sound is 340 m/s,
λ=
Diffraction minima occur at angles described by
a sin θ = mλ
(1.10 m ) sin θ1 = 1( 0.523 m )
θ1 = 28.4°
(1.10 m ) sin θ 2 = 2 ( 0.523 m )
θ 2 = 72.0°
(1.10 m ) sin θ3 = 3 ( 0.523 m )
θ3 nonexistent
Maxima appear straight ahead at 0°
and left and right at an angle given approximately by
(1.10 m ) sin θ x = 1.5 ( 0.523 m )
θ x ≈ 46°
There is no solution to a sin θ = 2.5λ , so our answer is already complete, with three sound
maxima.
Diffraction Patterns and Polarization
*P38.6
(a)
383
The rectangular patch on the wall is wider
than it is tall. The aperture will be taller
than it is wide. For horizontal spreading we
have
ywidth 0.110 m 2
=
= 0.012 2
L
4.5 m
awidth sin θ width = 1λ
tan θ width =
632.8 × 10 −9 m
= 5.18 × 10 −5 m
0.012 2
For vertical spreading, similarly
awidth =
0.006 m 2
= 0.000 667
4.5 m
632.8 × 10 −9 m
1λ
=
=
sin θ h
0.000 667
tan θ height =
aheight
FIG. P38.6
= 9.49 × 10 −4 m
(b) The central bright patch is horizontal. The aperture is vertical.
A smaller distance between aperture edges causes a wider diffraction angle. The longer dimension
of each rectangle is 18.3 times larger than the smaller dimension.
P38.7
sin θ ≈
y 4.10 × 10 −3 m
=
L
1.20 m
We define φ =
I
I max
P38.8
−4
π a sin θ π ( 4.00 × 10 m ) ⎛ 4.10 × 10 −3 m ⎞
=
⎟⎠ = 7.86 rad
λ
546.1 × 10 −9 m ⎜⎝
1.20 m
⎡ sin (φ ) ⎤
⎡ sin ( 7.86 ) ⎤
=⎢
=⎢
= 1.62 × 10 −2
⎥
⎣ 7.86 ⎥⎦
⎣ φ ⎦
2
2
mλ
, where m = ±1, ± 2, ± 3, . . . .
a
The requirement for m = 1 is from an analysis of the extra path distance
traveled by ray 1 compared to ray 3 in the textbook
λ
Figure 38.5. This extra distance must be equal to for destructive
2
interference. When the source rays approach the slit at an angle b, there is
a distance added to the path difference (of ray 1 compared to ray 3) of
a
sin β . Then, for destructive interference,
2
λ
λ
a
a
sin β + sin θ = so sin θ = 0 − sin β .
a
2
2
2
Equation 38.1 states that sin θ =
FIG. P38.8
Dividing the slit into 4 parts leads to the 2nd order minimum:
sin θ =
2λ
− sin β
a
Dividing the slit into 6 parts gives the third order minimum:
sin θ =
3λ
− sin β
a
Generalizing, we obtain the condition for the mth order minimum:
sin θ =
mλ
− sin β
a
384
*P38.9
Chapter 38
The diffraction envelope shows a broad central maximum
flanked by zeros at a sin θ = 1l and a sin θ = 2l. That is,
the zeros are at (π a sin θ )/λ = π , −π , 2π , −2π , . . . . Noting that the distance between slits is d = 9µm = 3a, we say
that within the diffraction envelope the interference pattern
shows closely spaced maxima at d sin θ = mλ , giving
(π 3a sin θ )/λ = mπ or (π a sin θ )/λ = 0, π /3, −π /3, 2π /3,
−2π /3. The third-order interference maxima are missing
because they fall at the same directions as diffraction
minima, but the fourth order can be visible at
(π a sin θ )/λ = 4π /3 and –4π /3 as diagrammed.
I
−p
0
p
p asinq
l
FIG. P38.9
P38.10
(a)
Double-slit interference maxima are at angles given
by d sin θ = mλ .
θ 0 = 0°
For m = 0,
For m = 1, ( 2.80 µ m ) sin θ = 1 ( 0.501 5 µ m ): θ1 = sin −1 ( 0.179 ) = 10.3°
Similarly, for m = 2, 3, 4, 5 and 6,
θ 2 = 21.0° , θ3 = 32.5° , θ 4 = 45.8°
θ5 = 63.6° , and θ6 = sin −1 (1.07 ) = nonexistent
Thus, there are 5 + 5 + 1 = 11 directions for interference maxima.
(b)
We check for missing orders by looking for single-slit diffraction minima, at a sin θ = mλ .
For m = 1,
and
θ1 = 45.8°
( 0.700 µm ) sin θ = 1( 0.501 5 µm )
Thus, there is no bright fringe at this angle. There are only nine bright fringes , at
θ = 0°, ± 10.3°, ± 21.0 °, ± 32.5°, and ± 63.6° .
(c)
⎡ sin (π a sin θ λ ) ⎤
I = I max ⎢
⎥
⎣ π sin θ λ
⎦
2
At θ = 0°,
sin θ
I
→ 1 and
→ 1.00
θ
I max
At θ = 10.3°,
π a sin θ π ( 0.700 µ m ) sin 10.3°
=
= 0.785 rad = 45.0°
0.501 5 µ m
λ
I
I max
sin 45.0° ⎤
= ⎡⎢
= 0.811
⎣ 0.785 ⎥⎦
2
Similarly, at θ = 21.0°,
π a sin θ
I
= 0.405
= 1.57 rad = 90.0° and
λ
I max
At θ = 32.5°,
π a sin θ
I
= 2.36 rad = 135° and
= 0.090 1
λ
I max
At θ = 63.6°,
π a sin θ
I
= 3.93 rad = 225° and
= 0.032 4
λ
I max
Diffraction Patterns and Polarization
Section 38.3
P38.11
385
Resolution of Single-Slit and Circular Apertures
We assume Rayleigh’s criterion applies to the predator’s eye with pupil narrowed. (It is a few
times too optimistic for a normal human eye with pupil dilated.)
sin θ =
λ 5.00 × 10 −7 m
=
= 1.00 × 10 −3 rad
a
5.00 × 10 −4
1.22λ /D = 1.22(589 nm) /(9 000 000 nm) = 79.8 µ rad
*P38.12 (a)
(b)
For a smaller angle of diffraction we choose the smallest visible wavelength, violet at
400 nm, to obtain 1.22λ / D = 1.22(400 nm) /(9 000 000 nm) = 54.2 µ rad .
(c)
The wavelength in water is shortened to its vacuum value divided by the index of refraction.
The resolving power is improved, with the miinimum resolvable angle becoming
1.22λ / D = 1.22(589 nm /1.33)/(9 000 000 nm) = 60.0 µ raad
Better than water for many purposes is oil immersion.
P38.13
Undergoing diffraction from a circular opening, the beam spreads into a cone of half-angle
⎛ 632.8 × 10 −9 m ⎞
λ
= 1.54 × 10 −4 rad
θ min = 1.22 = 1.22 ⎜
⎟
D
m
0
.
005
00
⎝
⎠
The radius of the beam ten kilometers away is, from the definition of radian measure,
rbeam = θ min (1.00 × 10 4 m ) = 1.544 m
and its diameter is dbeam = 2rbeam = 3.09 m .
*P38.14 When the pupil is open wide, it appears that the resolving power of human vision is limited by
the coarseness of light sensors on the retina. But we use Rayleigh’s criterion as a handy indicator
d
λ
of how good our vision might be. We take θ min = = 1.22 , where θ min is the smallest angular
L
D
separation of two objects for which they are resolved by an aperture of diameter D, d is the
separation of the two objects, and L is the maximum distance of the aperture from the two objects
at which they can be resolved.
Two objects can be resolved if their angular separation is greater than θ min. Thus, θ min should be
as small as possible. Therefore, light with the smaller of the two given wavelengths is easier to
resolve, i.e., blue .
L=
(5.20 × 10 −3 m ) ( 2.80 × 10 −2 m ) = 1.193 × 10 −4 m 2
Dd
=
1.22λ
1.22λ
λ
Thus L = 186 m for λ = 640 nm, and L = 271 m for λ = 440 nm. The viewer with the assumed
diffraction-limited vision could resolve adjacent tubes of blue in the range 186 m to 271 m , but
cannot resolve adjacent tubes of red in this range.
P38.15
When the pupil is open wide, it appears that the resolving power of human vision is limited by
the coarseness of light sensors on the retina. But we use Rayleigh’s criterion as a handy indicator
of how good our vision might be. According to this criterion, two dots separated center-to-center
by 2.00 mm would overlap
when
θ min =
Thus,
L=
d
λ
= 1.22
L
D
( 2.00 × 10 −3 m ) ( 4.00 × 10 −3 m ) = 13.1 m
dD
=
1.22λ
1.22 ( 500 × 10 −9 m )
386
P38.16
Chapter 38
When the pupil is open wide, it appears that the resolving power of human
vision is limited by the coarseness of light sensors on the retina. But we use
Rayleigh’s criterion as a handy indicator of how good our vision might be.
x = 1.22
P38.17
⎛ 5.00 × 10 −7 m ⎞
λ
D = 1.22 ⎜
( 250 × 103 m ) = 30.5 m
⎝ 5.00 × 10 −3 m ⎟⎠
d
D = 250 × 10 3 m
λ = 5.00 × 10 −7 m
d = 5.00 × 10 −3 m
The concave mirror of the spy satellite is probably about 2 m in diameter, and is surely not more
than 5 m in diameter. That is the size of the largest piece of glass successfully cast to a precise
shape, for the mirror of the Hale telescope on Mount Palomar. If the spy satellite had a larger mirror, its manufacture could not be kept secret, and it would be visible from the ground. Outer space
is probably closer than your state capitol, but the satellite is surely above 200-km altitude, for
reasonably low air friction. We find the distance between barely resolvable objects at a distance
of 200 km, seen in yellow light through a 5-m aperture:
y
λ
= 1.22
L
D
⎛ 6 × 10 −7 m ⎞
y = ( 200 000 m ) (1.22 ) ⎜
= 3 cm
⎝
5 m ⎠⎟
(Considering atmospheric seeing caused by variations in air density and temperature, the distance
⎛ 1° ⎞ ⎛ π rad ⎞
between barely resolvable objects is more like ( 200 000 m ) (1 s ) ⎜
⎜
⎟ = 97 cm.)
⎝ 3 600 s ⎟⎠ ⎝ 180 ° ⎠
Thus the snooping spy satellite cannot see the difference between III and II or IV on a license
plate. It cannot count coins spilled on a sidewalk, much less read the dates on them.
P38.18
1.22
λ d
=
D L
D = 2.10 m
d = 1.22
Section 38.4
P38.19
d=
λ=
c
= 0.020 0 m
f
L = 9 000 m
( 0.020 0 m ) ( 9 000 m )
2.10 m
= 105 m
The Diffraction Grating
1.00 cm 1.00 × 10 −2 m
=
= 5.00 µ m
2 000
2 000
sin θ =
−9
mλ 1 ( 640 × 10 m )
=
= 0.128 θ = 7.35°
d
5.00 × 10 −6 m
Diffraction Patterns and Polarization
P38.20
The principal maxima are defined by
d sin θ = mλ
For m = 1,
m = 0, 1, 2, . . .
λ = d sin θ
q
where q is the angle between the central (m = 0) and the first
order (m = 1) maxima. The value of q can be determined from
the information given about the distance between maxima and the
grating-to-screen distance. From the figure,
tan θ =
so
and
0.488 m
= 0.284
1.72 m
0.488 m
1.72 m
FIG. P38.20
θ = 15.8°
sin θ = 0.273
The distance between grating “slits” equals the reciprocal of the number of grating lines per
centimeter
d=
The wavelength is
P38.21
387
1
= 1.88 × 10 −4 cm = 1.88 × 10 3 nm
5 310 cm −1
λ = d sin θ = (1.88 × 10 3 nm ) ( 0.273) = 514 nm
The grating spacing is
d=
1.00 × 10 −2 m
= 2.22 × 10 −6 m
4 500
In the 1st-order spectrum, diffraction angles are given by
sin θ =
λ
:
d
sin θ1 =
656 × 10 −9 m
= 0.295
2.22 × 10 −6 m
so that for red
θ1 = 17.17°
and for blue
sin θ 2 =
so that
θ 2 = 11.26°
FIG. P38.21
434 × 10 −9 m
= 0.195
2.22 × 10 −6 m
The angular separation is in first-order,
∆θ = 17.17° − 11.26° = 5.91°
In the second-order spectrum,
⎛ 2λ ⎞
⎛ 2λ ⎞
∆θ = sin −1 ⎜ 1 ⎟ − sin −1 ⎜ 2 ⎟ = 13.2°
⎝ d ⎠
⎝ d ⎠
Again, in the third order,
⎛ 3λ ⎞
⎛ 3λ ⎞
∆θ = sin −1 ⎜ 1 ⎟ − sin −1 ⎜ 2 ⎟ = 26.5°
⎝ d ⎠
⎝ d ⎠
Since the red does not appear in the fourth-order spectrum, the answer is complete.
388
P38.22
Chapter 38
sin θ = 0.350:
d=
λ
632.8 nm
=
= 1.81 × 10 3 nm
sin θ
0.350
Line spacing = 1.81 µ m
P38.23
(a)
(b)
d=
1
= 2.732 × 10 −4 cm = 2.732 × 10 −6 m = 2 732 nm
3 660 lines cm
λ=
d sin θ
:
m
d=
λ
sin θ1
At θ = 10.09°,
λ = 478.7 nm
At θ = 13.71°,
λ = 647.6 nm
At θ = 14.77°,
λ = 696.6 nm
and
2λ = d sin θ 2
so
sin θ 2 =
2λ
2λ
=
= 2 sin θ1
d
λ sin θ1
Therefore, if θ1 = 10.09° then sin θ 2 = 2 sin(10.09°) gives θ 2 =
Similarly, for θ1 = 13.71°, θ 2 = 28.30°
P38.24
sin θ =
20.51° .
and for θ1 = 14.77°, θ 2 = 30.66° .
mλ
d
Therefore, taking the ends of the visible spectrum to be λ v = 400 nm and λr = 750 nm, the ends
of the different order spectra are defined by:
2λr 1 500 nm
=
d
d
3λ
1 200 nm
= v =
d
d
End of second order:
sin θ 2 r =
Start of third order:
sin θ3 v
Thus, it is seen that θ 2 r > θ3 v and these orders must overlap regardless of the value of the
grating spacing d.
*P38.25 The sound has wavelength λ =
by
v
343 m s
=
= 9.22 × 10 −3 m. Each diffracted beam is described
f 37.2 × 10 3 s
d sinθ = mλ , m = 0, 1, 2, . . .
The zero-order beam is at m = 0, θ = 0. The beams in the first order of interference are to the
1λ
9.22 × 10 −3 m
= sin −1
= sin −1 0.709 = 45.2°. For a second-order beam
d
1.3 × 10 −2 m
2λ
= sin −1 2 ( 0.709 ) = sin −1 1.42. No angle, smaller or larger than 90°,
we would need θ = sin −1
d
left and right at θ = sin −1
has a sine greater than 1. Then a diffracted beam does not exist for the second order or any higher
order. The whole answer is then, three beams, at 0 ° and at 45.2° to the right and to the left .
Diffraction Patterns and Polarization
P38.26
y 0.4 µ m
=
L 6.9 µ m
θ = 3.32°
For a side maximum, tan θ =
d sin θ = mλ
d=
(1) ( 780 × 10 −9 m )
sin 3.32°
The number of grooves per millimeter =
= 13.5 µ m
1 × 10 −3 m
13.5 × 10 −6 m
= 74.2
P38.27
d=
(a)
1.00 × 10 −3 m mm
= 4.00 × 10 −6 m = 4 000 nm
250 lines mm
or
(b)
d sin θ = mλ ⇒ m =
d sin θ
λ
d sin θ max ( 4 000 nm ) sin 90.0°
=
= 5.71
λ
700 nm
5 orders is the maximum
The highest order in which the violet end of the spectrum can be seen is:
mmax =
or
(a)
FIG. P38.26
The number of times a complete order is seen is the same as the number of orders in which
the long wavelength limit is visible.
mmax =
P38.28
389
d sin θ max ( 4 000 nm ) sin 90.0°
=
= 10.0
400 nm
λ
10 orders in the short-wavelength region
The several narrow parallel slits make a diffraction
grating. The zeroth- and first-order maxima are separated
according to
λ 632.8 × 10 −9 m
d sin θ = (1) λ
sin θ = =
d
1.2 × 10 −3 m
θ = sin −1 ( 0.000 527 ) = 0.000 527 rad
y = L tan θ = (1.40 m ) ( 0.000 527 ) = 0.738 mm
(b)
Many equally spaced transparent lines appear on the
film. It is itself a diffraction grating. When the same
light is sent through the film, it produces interference
maxima separated according to
FIG. P38.28
λ 632.8 × 10 −9 m
= 0.000 857
=
d 0.738 × 10 −3 m
y = L tan θ = (1.40 m ) ( 0.000 857 ) = 1.20 mm
d sin θ = (1) λ
sin θ =
An image of the original set of slits appears on the screen. If the screen is removed, light
diverges from the real images with the same wave fronts reconstructed as the original
slits produced. Reasoning from the mathematics of Fourier transforms, Gabor showed
that light diverging from any object, not just a set of slits, could be used. In the picture,
the slits or maxima on the left are separated by 1.20 mm. The slits or maxima on the right
are separated by 0.738 mm. The length difference between any pair of lines is an integer
number of wavelengths. Light can be sent through equally well toward the right or toward
the left. Soccer players shift smoothly between offensive and defensive tactics.
390
P38.29
Chapter 38
1
= 2.38 × 10 −6 m = 2 380 nm
4 200 cm
⎛ mλ ⎞
d sin θ = mλ or
θ = sin −1 ⎜
and
⎝ d ⎟⎠
d=
⎡
⎛ mλ ⎞ ⎤
y = L tan θ = L tan ⎢sin −1 ⎜
⎝ d ⎟⎠ ⎥⎦
⎣
Thus,
⎧ ⎡
⎡
⎛ mλ 2 ⎞ ⎤
⎛ mλ ⎞ ⎤ ⎫
∆y = L ⎨ tan ⎢sin −1 ⎜
− tan ⎢sin −1 ⎜ 1 ⎟ ⎥ ⎬
⎟
⎥
⎝ d ⎠⎦
⎝ d ⎠ ⎦⎭
⎣
⎩ ⎣
For m = 1,
⎧⎪ ⎡
⎡ −1 ⎛ 589 ⎞ ⎤ ⎫⎪
⎛ 589.6 ⎞ ⎤
∆y = ( 2.00 m ) ⎨ tan ⎢sin −1 ⎜
⎥ − tan ⎢sin ⎜
⎥ ⎬ = 0.554 mm
⎟
⎝ 2 380 ⎠ ⎦
⎝ 2 380 ⎟⎠ ⎦ ⎭⎪
⎣
⎩⎪ ⎣
For m = 2,
⎧⎪ ⎡
⎡ −1 ⎛ 2 ( 589 ) ⎞ ⎤ ⎫⎪
⎛ 2 ( 589.6 ) ⎞ ⎤
∆y = ( 2.00 m ) ⎨ tan ⎢sin −1 ⎜
⎥ − tan ⎢sin ⎜
⎥ ⎬ = 1.54 mm
⎟
⎝ 2 380 ⎠ ⎦
⎝ 2 380 ⎟⎠ ⎦ ⎭⎪
⎣
⎩⎪ ⎣
For m = 3,
⎧⎪ ⎡
⎡ −1 ⎛ 3 ( 589 ) ⎞ ⎤ ⎫⎪
⎛ 3 ( 589.6 ) ⎞ ⎤
∆y = ( 2.00 m ) ⎨ tan ⎢sin −1 ⎜
⎥ − tan ⎢sin ⎜
⎥ ⎬ = 5.04 mm
⎟
⎝ 2 380 ⎠ ⎦
⎝ 2 380 ⎟⎠ ⎦ ⎭⎪
⎣
⎩⎪ ⎣
Thus, the observed order must be m = 2 .
Section 38.5
Diffraction of X-Rays by Crystals
−9
2d sin θ 2 ( 0.353 × 10 m ) sin 7.60°
= 9.34 × 10 −11 m = 0.093 4 nm
=
m
1
P38.30
2d sin θ = mλ :
λ=
P38.31
2d sin θ = mλ :
sin θ =
and
P38.32
−9
mλ 1 ( 0.140 × 10 m )
=
= 0.249
2d 2 ( 0.281 × 10 −9 m )
θ = 14.4°
Figure 38.23 of the text shows the situation.
2d sin θ = mλ
m = 1:
m = 2:
m = 3:
or
2d sin θ
m
2 ( 2.80 m ) sin 80.0°
= 5.51 m
λ1 =
1
2 ( 2.80 m ) sin 80.0°
λ2 =
= 2.76 m
2
2 ( 2.80 m ) sin 80.0°
= 1.84 m
λ3 =
3
λ=
*P38.33 The crystal cannot produce diffracted beams of visible light. The wavelengths of visible light are
some hundreds of nanometers. There is no angle whose sine is greater than 1. Bragg’s law
2d sin q = ml cannot be satisfied for a wavelength much larger than the distance between atomic
planes in the crystal.
Diffraction Patterns and Polarization
Section 38.6
P38.34
391
Polarization of Light Waves
The average value of the cosine-squared function is one-half, so the first polarizer transmits
3
1
the light. The second transmits cos 2 30.0° = .
4
2
1 3
3
I f = × Ii =
Ii
2 4
8
P38.35
I = I max cos 2 θ
θ = cos −1
⇒
I
I max
(a)
I
1
=
I max 3.00
⇒
θ = cos −1
1
= 54.7°
3.00
(b)
I
1
=
I max 5.00
⇒
θ = cos −1
1
= 63.4°
5.00
(c)
I
1
=
I max 10.0
⇒
θ = cos −1
1
= 71.6°
10.0
P38.36
By Brewster’s law,
P38.37
sin θ c =
1
n
Also, tan θ p = n.
n = tan θ p = tan ( 48.0° ) = 1.11
1
1
=
= 1.77
sin θ c sin 34.4°
or
n=
Thus,
θ p = tan −1 ( n ) = tan −1 (1.77 ) = 60.5°
1
and tan θ p = n
n
1
Thus, sin θ c =
or cot θ p = sin θ c .
tan θ p
P38.38
sin θ c =
P38.39
(a)
At incidence, n1 sin θ1 = n2 sin θ 2 and θ1′ = θ1. For complete
polarization of the reflected light,
(90 − θ1′ ) + (90 − θ 2 ) = 90°
q 1 q 1'
n1
n2
θ1′ + θ 2 = 90 = θ1 + θ 2
Then n1 sin θ1 = n2 sin ( 90 − θ1 ) = n2 cos θ1
sin θ1 n2
=
= tan θ1
cos θ1 n1
q2
n1
At the bottom surface, θ 3 = θ 2 because the normals to the
surfaces of entry and exit are parallel.
Then
q 3 q 3'
n2 sin θ 3 = n1 sin θ 4
and
θ 3′ = θ 3
n2 sin θ 2 = n1 sin θ 4
and
θ 4 = θ1
q4
FIG. P38.39(a)
The condition for complete polarization of the reflected light is
90 − θ 3′ + 90 − θ 4 = 90°
θ 2 + θ1 = 90°
This is the same as the condition for θ1 to be Brewster’s angle at the top surface.
continued on next page
392
Chapter 38
(b)
We consider light moving in a plane perpendicular to
the line where the surfaces of the prism meet at the
unknown angle Φ. We require
Φ
n1 sin θ1 = n2 sin θ 2
n1
n2
q1 q1
n3
q2
θ1 + θ 2 = 90°
So n1 sin ( 90 − θ 2 ) = n2 sin θ 2
And
n2 sin θ 3 = n3 sin θ 4
n1
= tan θ 2
n2
q4
θ 3 + θ 4 = 90°
FIG. P38.39(b)
n
n2 sin θ 3 = n3 cos θ 3
tan θ 3 = 3
n2
In the triangle made by the faces of the prism and the ray in the prism,
Φ + 90 + θ 2 + ( 90 − θ3 ) = 180
So one particular apex angle is required, annd it is
⎛n ⎞
⎛n ⎞
Φ = θ 3 − θ 2 = tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 1 ⎟
⎝ n2 ⎠
⎝ n2 ⎠
Here a negative result is to be interpreted as meaning the same as a positive result.
P38.40
For incident unpolarized light of intensity I max:
The average value of the cosine-squared function is one
half, so the intensity after transmission by the first disk is
I=
1
I max
2
After transmitting 2nd disk: I =
After transmitting 3rd disk:
1
I max cos 2 θ
2
I=
FIG. P38.40
1
I max cos 2 θ cos 2 ( 90° − θ )
2
where the angle between the first and second disk is θ = ω t .
Using trigonometric identities cos 2 θ =
1
(1 + cos 2θ )
2
and
cos 2 ( 90° − θ ) = sin 2 θ =
we have
I=
q3
1
(1 − cos 2θ )
2
1
⎡ (1 + cos 2θ ) ⎤ ⎡ (1 − cos 2θ ) ⎤
I max ⎢
⎥⎦ ⎢⎣
⎥⎦
2
2
2
⎣
1
1
1
I = I max (1 − cos 2 2θ ) = I max ⎛ ⎞ (1 − cos 4θ )
⎝
8
8
2⎠
Since θ = ω t, the intensity of the emerging beam is given by I =
1
I max (1 − 4ω t ) .
16
q3
Diffraction Patterns and Polarization
*P38.41 (a)
393
Let I0 represent the intensity of unpolarized light incident on the first polarizer. In Malus’s
law the average value of the cosine-squared function is 1/2, so the first filter lets through 1/2
of the incident intensity. Of the light reaching them, the second filter passes cos245° = 1/2 and
the third filter also cos245° = 1/2. The transmitted intensity is then I0(1/2)(1/2)(1/2) = 0.125 I0.
The reduction in intensity is by a factor of 0.875 of the incident intensity.
(b)
By the same logic as in part (a) we have transmitted I0(1/2)(cos230°)(cos230°)(cos230°) =
I0(1/2)(cos230°)3 = 0.211 I0. Then the fraction absorbed is 0.789 .
(c)
Yet again we compute transmission I0(1/2)(cos215°)6 = 0.330 I0. And the fraction absorbed
is 0.670 .
(d)
We can get more and more of the incident light through the stack of ideal filters, approaching 50%, by reducing the angle between the transmission axes of each one and the next.
Additional Problems
*P38.42 The central bright fringe is wider than the side bright fringes, so the light must have been
diffracted by a single slit . For precision, we measure from the second minimum on one side
of the center to the second minimum on the other side:
2 y = (11.7 − 6.3) cm = 5.4 cm
y = 2.7 cm
y 0.027 m
tan θ = =
= 0.010 4
L
2.6 m
θ = 0.595° = 0.010 4 rad
a sin θ = mλ
a=
P38.43
−9
−9
mλ 2 ( 632.8 × 10 m ) 2 ( 632.8 × 10 m )
=
=
= 1.22 × 10 −4 m
sin 0.595°
0.010 4
sin θ
Let the first sheet have its axis at angle q to the original plane of polarization, and let each further
sheet have its axis turned by the same angle.
The first sheet passes intensity
I max cos 2 θ
The second sheet passes
I max cos 4 θ
and the nth sheet lets through
I max cos 2 n θ ≥ 0.90 I max
Try different integers to find
45° ⎞
cos 2×5 ⎛⎜
= 0.885
⎝ 5 ⎟⎠
(a)
So n = 6
(b)
θ = 7.50°
45°
n
45° ⎞
cos 2×6 ⎛⎜
= 0.902
⎝ 6 ⎟⎠
where θ =
394
P38.44
Chapter 38
Consider vocal sound moving at 340 m/s and of frequency 3 000 Hz. Its wavelength is
λ=
v 340 m s
=
= 0.113 m
f 3 000 Hz
If your mouth, for horizontal dispersion, behaves similarly to a slit 6.00 cm wide, then
a sin θ = mλ predicts no diffraction minima. You are a nearly isotropic source of this sound. It
spreads out from you nearly equally in all directions. On the other hand, if you use a megaphone
with width 60.0 cm at its wide end, then a sin θ = mλ predicts the first diffraction minimum at
0.113 m ⎞
⎛ mλ ⎞
θ = sin −1 ⎜
= sin −1 ⎛⎜
= 10.9°
⎝ 0.600 m ⎟⎠
⎝ a ⎟⎠
This suggests that the sound is radiated mostly toward the front into a diverging beam of angular
diameter only about 20°. With less sound energy wasted in other directions, more is available for
your intended auditors. We could check that a distant observer to the side or behind you receives
less sound when a megaphone is used.
*P38.45 (a)
We assume the first side maximum is at a sinq = 1.5 l. (Its location is determined more
precisely in problem 66.) Then the required fractional intensity is
I
I max
(b)
2
1
⎡ sin(π a sin θ /λ ) ⎤
⎡ sin(1.5π ) ⎤
=⎢
⎥ = ⎢⎣ 1.5π ⎦⎥ = 2.25π 2 = 0.0450
π
a
sin
θ
/
λ
⎣
⎦
2
Proceeding as in part (a) we have a sinq/l = 2.5 and
I
I max
2
sin(2.5π ) ⎤
1
⎡ sin(π a sin θ /λ ) ⎤
=
= 0.0162
=⎢
= ⎡⎢
⎥
⎥
6.25π 2
⎣ 2.5π ⎦
⎣ π a sin θ /λ ⎦
2
*P38.46 The energy in the central maximum we can estimate in Figure 38.6 as proportional to
(height)(width) = Imax(2p)
As in problem 45, the maximum height of the first side maximum is approximately
Imax[sin (3p /2)/(3p /2)]2 = 4Imax/9p 2
Then the energy in one side maximum is proportional to p (4Imax/9p 2),
and that in both of the first side maxima together is proportional to 2p (4Imax/9p 2).
Similarly and more precisely, and always with the same proportionality constant, the energy in
both of the second side maxima is proportional to 2p (4Imax/25p 2).
The energy in all of the side maxima together is proportional to
2p (4Imax/p 2)[1/32 + 1/52 + 1/72 + …] = 2p (4Imax/p 2)[p 2/8 – 1] = (8Imax/p )[p 2/8 – 1]
= Imax(p – 8/p) = 0.595Imax
The ratio of the energy in the central maximum to the total energy is then
Imax(2p)/{Imax(2p) + 0.595Imax} = 2p /6.88 = 0.913 = 91.3%
Our calculation is only a rough estimate, because the shape of the central maximum in particular
is not just a vertically-stretched cycle of a cosine curve. It is slimmer than that.
Diffraction Patterns and Polarization
P38.47
The first minimum is at
a sin θ = (1) λ
This has no solution if
λ
>1
a
or if
a < λ = 632.8 nm
395
*P38.48 With light in effect moving through vacuum, Rayleigh’s criterion limits the resolution according to
D = 1.22 lL/d = 1.22 (885 × 10–9 m) 12 000 m/2.3 m = 5.63 mm
d/L = 1.22 l/D
The assumption is absurd. Over a horizontal path of 12 km in air, density variations associated
with convection (“heat waves” or what an astronomer calls “seeing”) would make the motorcycles completely unresolvable with any optical device.
P38.49
d=
1
= 2.50 × 10 −6 m
400 mm −1
(a)
d sin θ = mλ
(b)
λ=
(c)
d sin θ a = 2λ
⎛ 2 × 541 × 10 −9 m ⎞
θ a = sin −1 ⎜
= 25.6°
⎝ 2.50 × 10 −6 m ⎟⎠
541 × 10 −9 m
= 4.07 × 10 −7 m
1.33
and
d sin θ b =
2λ
n
combine by substitution to give
P38.50
(a)
λ=
v
:
f
θ min = 1.22
λ=
λ
:
D
n sin θb = (1) sin θ a
3.00 × 108 m s
= 0.214 m
1.40 × 10 9 s −1
0.214 m ⎞
θ min = 1.22 ⎛
= 7.26 µ rad
⎝ 3.60 × 10 4 m ⎠
θ min = 7.26 µ rad ⎛
⎝
d
:
L
⎛ 2 × 4.07 × 10 −7 m ⎞
θb = sin −1 ⎜
= 19.0°
⎝ 2.50 × 10 −6 m ⎟⎠
180 × 60 × 60 s ⎞
= 1.50 arc secoonds
⎠
π
d = θ min L = ( 7.26 × 10 −6 rad ) ( 26 000 ly ) = 0.189 ly
(b)
θ min =
(c)
It is not true for humans, but we assume the hawk’s visual acuity is limited only by
Rayleigh’s criterion.
θ min = 1.22
(d)
⎛ 500 × 10 −9 m ⎞
λ
θ min = 1.22 ⎜
= 50.8 µ rad (10.5 seconds of arc )
⎝ 12.0 × 10 −3 m ⎟⎠
D
d = θ min L = ( 50.8 × 10 −6 rad ) ( 30.0 m ) = 1.52 × 10 −3 m = 1.52 mm
396
P38.51
Chapter 38
With a grazing angle of 36.0°, the angle of incidence is 54.0°
tan θ p = n = tan 54.0° = 1.38
In the liquid, λn =
P38.52
(a)
λ 750 nm
=
= 545 nm .
n
1.38
Bragg’s law applies to the space lattice of melanin rods. Consider the planes d = 0.25 µ m
apart. For light at near-normal incidence, strong reflection happens for the wavelength
given by 2d sin θ = mλ . The longest wavelength reflected strongly corresponds to m = 1:
2 ( 0.25 × 10 −6 m ) sin 90° = 1λ
(b)
λ = 500 nm. This is the blue-green color.
For light incident at grazing angle 60°, 2d sin θ = mλ
gives 1λ = 2 ( 0.25 × 10 −6 m ) sin 60° = 433 nm. This is violet.
P38.53
(c)
Your two eyes receive light reflected from the feather at different angles, so they receive
light incident at different angles and containing different colors reinforced by constructive
interference.
(d)
The longest wavelength that can be reflected with extra strength by these melanin rods is
the one we computed first, 500 nm blue-green.
(e)
If the melanin rods were farther apart (say 0.32 µ m) they could reflect red with
constructive interference.
(a)
d sin θ = mλ
−9
mλ 3 ( 500 × 10 m )
=
= 2.83 µ m
sin θ
sin 32.0°
1
1
Therefore, lines per unit length = =
d 2.83 × 10 −6 m
or d =
or lines per unit length = 3.53 × 10 5 m −1 = 3.53 × 10 3 cm −1 .
(b)
sin θ =
−9
mλ m ( 500 × 10 m )
=
= m ( 0.177 )
−6
d
2.83 × 10 m
For sin θ ≤ 1.00, we must have
m ( 0.177 ) ≤ 1.00
or
m ≤ 5.66
Therefore, the highest order observed is
m=5
Total number of primary maxima observed is
2 m + 1 = 11
Diffraction Patterns and Polarization
*P38.54 (a)
For the air-to-water interface,
n
1.33
tan θ p = water =
1.00
nair
and
qp
θ p = 53.1°
q
(1.00 ) sin θ p = (1.33) sin θ 2
θ 2 = sin −1 ⎛
⎝
397
q2
q3
Air
Water
sin 53.1° ⎞
= 36.9°
1.33 ⎠
For the water-to-slab interface, tan θ p = tan θ 3 =
nslab
n
=
nwater 1.33
FIG. P38.54
so θ3 = tan −1 (n/1.33).
The angle between surfaces is θ = θ3 − θ 2 = tan−1(n /1.33) − 36.9°.
P38.55
(b)
If we imagine n → ∞, then θ → 53.1°. The material of the slab in this case is higher in optical density than any gem. The light in the water skims along the upper surface of the slab.
(c)
If we imagine n = 1, then θ = 0. The slab is so low in optical density that it is like air. The
light strikes parallel surfaces as it enters and exits the water, both at the polarizing angle.
A central maximum and side maxima in seven orders of interference appear.
If the seventh order is just at 90°,
d sin θ = mλ
d (1) = 7 ( 654 × 10 −9 m )
d = 4.58 µ m
If the seventh order is at less than 90°, the eighth order might be nearly
ready to appear according to
d (1) = 8 ( 654 × 10 −9 m )
d = 5.23 µ m
FIG. P38.55
Thus 4.58 µ m < d < 5.23 µ m .
P38.56
(a)
We require
Then
(b)
P38.57
θ min = 1.22
λ radius of diffraction disk D
=
=
D
L
2L
D 2 = 2.44 λ L
D = 2.44 ( 500 × 10 −9 m ) ( 0.150 m ) = 428 µ m
For the limiting angle of resolution between lines we
assume θ min = 1.22
(550 × 10 −9 m ) = 1.34 × 10 −4 rad.
λ
= 1.22
D
(5.00 × 10 −3 m )
Assuming a picture screen with vertical dimension , the minimum viewing distance for no
485
visible lines is found from θ min =
. The desired ratio is then
L
L
1
1
= 15.4
=
=
485θ min 485 (1.34 × 10 −4 rad )
When the pupil of a human eye is wide open, its actual resolving power is significantly poorer
than Rayleigh’s criterion suggests.
398
P38.58
Chapter 38
(a)
Applying Snell’s law gives n2 sin φ = n1 sin θ . From the sketch,
we also see that:
θ + φ + β = π , or φ = π − (θ + β )
FIG. P38.58(a)
Using the given identity:
sin φ = sin π cos (θ + β ) − cos π sin (θ + β )
which reduces to:
sin φ = sin (θ + β )
Applying the identity again:
sin φ = sin θ cos β + cos θ sin β
Snell’s law then becomes:
n2 ( sin θ cos β + cos θ sin β ) = n1 sin θ
or (after dividing by cosq ):
n2 ( tan θ cos β + sin β ) = n1 tan θ
tan θ =
Solving for tan θ gives:
(b)
If β = 90.0°, n1 = 1.00, and n2 = n, the above result becomes:
tan θ =
P38.59
(a)
(b)
n (1.00 )
, or n = tan θ , which is Brewster’s law.
1.00 − 0
From Equation 38.1,
⎛ mλ ⎞
θ = sin −1 ⎜
⎝ a ⎟⎠
In this case m = 1 and
λ=
Thus,
⎛ 4.00 × 10 −2 m ⎞
θ = sin −1 ⎜
= 41.8°
⎝ 6.00 × 10 −2 m ⎟⎠
From Equation 38.2,
When θ = 15.0°,
sin θ =
I
I max
φ=
c 3.00 × 108 m s
=
= 4.00 × 10 −2 m
f
7.50 × 10 9 Hz
⎡ sin (φ ) ⎤
=⎢
⎥
⎣ φ ⎦
2
where φ =
π a sin θ
λ
π ( 0.060 0 m ) sin 15.0°
= 1.22 rad
0.040 0 m
⎡ sin (1.22 rad ) ⎤
=⎢
= 0.593
⎣ 1.22 rad ⎥⎦
2
I
and
(c)
n2 sin β
n1 − n2 cos β
I max
λ
so θ = 41.8°:
a
This is the minimum angle subtended by the two sources at the slit. Let
a be the half angle between the sources, each a distance = 0.100 m
from the center line and a distance L from the slit plane. Then,
L = cot α = ( 0.100 m ) cot ⎛
⎝
41.8° ⎞
= 0.262 m
2 ⎠
FIG. P38.59(c)
Diffraction Patterns and Polarization
*P38.60 (a)
The first sheet transmits one-half the intensity of the originally unpolarized light, because
the average value of the cosine-squared function in Malus’s law is one-half. Then
(
1
I
= cos 2 45.0°)
I max 2
P38.61
399
( cos
2
45.0 °) =
1
8
(b)
No recipes remain. The two experiments follow precisely analogous steps, but the results
are different. The middle filter in part (a) changes the polarization state of the light that
passes through it, but the recipe selections do not change individual recipes. The result for
light gives us a glimpse of how quantum-mechanical measurements differ from classical
measurements.
(a)
Constructive interference of light of wavelength l on the screen is described by
d sin θ = mλ where tan θ =
y
so sin θ =
L
y
L +y
2
2
. Then ( d ) y ( L2 + y 2 )
−1 2
= mλ .
Differentiating with respect to y gives
d1 ( L2 + y 2 )
d
(L
2
+y
)
2 12
−1 2
−
−3 2
1
dλ
+ ( d ) y ⎛ − ⎞ ( L2 + y 2 ) ( 0 + 2 y ) = m
⎝ 2⎠
dy
(L
( d ) y2
2
+y
)
2 32
=m
d λ ( d ) L2 + ( d ) y 2 − ( d ) y 2
=
dy
( L2 + y2 )3 2
dλ
(d ) L
=
32
dy m ( L2 + y 2 )
2
(b)
Here d sin θ = mλ gives
10 −2 m
⎛ 0.55 × 10 −6 m ⎞
sin θ = 1 ( 550 × 10 −9 m ), θ = sin −1 ⎜
= 26.1°
8 000
⎝ 1.25 × 10 −6 m ⎟⎠
y = L tan θ = 2.40 m tan 26.1° = 1.18 m
dλ
dL2
1.25 × 10 −6 m ( 2.40 m )
=
=
= 3.77 × 10 −7 = 3.77 nm cm .
3
2
2
2 32
dy m ( L2 + y 2 )
1 ( 2.4 m ) + (1.18 m )
2
Now
P38.62
(a)
The angles of bright beams diffracted from the grating are given by ( d ) sin θ = mλ . The
angular dispersion is defined as the derivative
(b)
)
(
dθ
:
dλ
( d ) cos θ
dθ
=m
dλ
For the average wavelength 578 nm,
0.02 m
sin θ = 2 ( 578 × 10 −9 m )
8 000
2 × 578 × 10 −9 m
= 27.5°
θ = sin −1
2.5 × 10 −6 m
d sin θ = mλ
The separation angle between the lines is
2
dθ
m
2.11 × 10 −9 m
∆λ =
∆λ =
2.5 × 10 −6 m cos 27.5°
dλ
d cos θ
180° ⎞
= 0.109°
= 0.001 90 = 0.001 90 rad = 0.001 90 rad ⎛
⎝ π rad ⎠
∆θ =
dθ
m
=
d λ d cos θ
400
P38.63
Chapter 38
(a)
The E and O rays, in phase at the surface of the plate, will have a phase difference
2π
θ = ⎛ ⎞δ
⎝ λ ⎠
after traveling distance d through the plate. Here d is the difference in the optical path
lengths of these rays. The optical path length between two points is the product of the
actual path length d and the index of refraction. Therefore,
δ = dnO − dnE
The absolute value is used since nO may be more or less than unity. Therefore,
nE
2π
2π
θ = ⎛ ⎞ dnO − dnE = ⎛ ⎞ d nO − nE
⎝ λ ⎠
⎝ λ ⎠
(b)
P38.64
(a)
d=
(550 × 10 −9 m ) (π 2) = 1.53 × 10 −5 m = 15.3 µm
λθ
=
2π nO − nE
2π 1.544 − 1.553
From Equation 38.2,
where we define
Therefore, when
(b)
I
I max
φ≡
I
I max
Let y1 = sin φ and y2 =
⎡ sin (φ ) ⎤
=⎢
⎥
⎣ φ ⎦
2
π a sin θ
λ
=
1
2
we must have
sin φ
φ
1
, or sin φ =
=
φ
2
2
φ
.
2
A plot of y1 and y2 in the range 1.00 ≤ φ ≤
π
is shown to
2
the right.
The solution to the transcendental equation is found to
be φ = 1.39 rad .
(c)
π a sin θ
λ
φ λ
= φ gives sin θ = ⎛ ⎞ = 0.443 .
⎝
⎠
λ
π a
a
FIG. P38.64(b)
λ
If λ is small, then θ ≈ 0.443 .
a
a
This gives the half-width, measured away from the maximum at θ = 0. The pattern is
symmetric, so the full width is given by
∆θ = 0.443
0.886λ
λ ⎛
λ⎞
− ⎜ −0.443 ⎟ =
a ⎝
a⎠
a
Diffraction Patterns and Polarization
P38.65
2 sin φ
f
1
1.19
bigger than f
2
1.29
smaller than f
1.5
1.41
smaller
1.4
1.394
1.39
1.391
1.395
1.392
1.392
1.391 7
smaller
1.391 5
1.391 54
bigger
1.391 52
1.391 55
bigger
1.391 6
1.391 568
smaller
1.391 58
1.391 563
1.391 57
1.391 561
1.391 56
1.391 558
1.391 559
1.391 557 8
1.391 558
1.391 557 5
1.391 557
1.391 557 3
1.391 557 4
1.391 557 4
bigger
We get the answer as 1.391 557 4 to seven digits after 17 steps. Clever guessing, like using the
value of 2 sin φ as the next guess for f, could reduce this to around 13 steps.
⎛ 2 sin (φ ) ⎞ ⎡ (φ ) cos (φ ) − sin (φ ) ⎤
⎡ sin (φ ) ⎤
dI
In I = I max ⎢
we find
= I max ⎜
⎢
⎥
⎥
dφ
⎝ φ ⎟⎠ ⎢⎣
( φ )2
⎥⎦
⎣ φ ⎦
2
P38.66
and require that it be zero. The possibility sin (φ ) = 0 locates all of the minima and the central
maximum, according to
π a sin θ
= 0, π , 2π , . . . ; a sin θ = 0, λ , 2λ , . . .
λ
φ = 0, π , 2π , . . .;
φ=
The side maxima are found from
φ cos (φ ) − sin (φ ) = 0, or tan (φ ) = φ
This has solutions
φ = 4.493 4 , φ = 7.725 3 , and others, giving
(a)
π a sin θ = 4.493 4 λ
a sin θ = 1.430 3λ
(b)
π a sin θ = 7.725 3λ
a sin θ = 2.459 0 λ
401
402
P38.67
Chapter 38
The first minimum in the single-slit diffraction
pattern occurs at
sin θ =
λ ymin
≈
a
L
Thus, the slit width is given by
a=
λL
ymin
For a minimum located at ymin = 6.36 mm ± 0.08 mm,
the width is
(632.8 × 10
a=
−9
m ) (1.00 m )
6.36 × 10 −3 m
FIG. P38.67
= 99.5 µ m ± 1%
ANSWERS TO EVEN PROBLEMS
P38.2
547 nm
P38.4
91.2 cm
P38.6
(a) 51.8 mm wide and 949 mm high (b) horizontal; vertical. See the solution. A smaller distance
between aperture edges causes a wider diffraction angle. The longer dimension of each rectangle
is 18.3 times larger than the smaller dimension.
P38.8
See the solution.
P38.10
(a) 0°, 10.3°, 21.0°, 32.5°, 45.8°, 63.6° (b) nine bright fringes, at 0° and on either side at 10.3°,
21.0°, 32.5°, and 63.6° (c) 1.00, 0.811, 0.405, 0.090 1, 0.032 4
P38.12
(a) 79.8 mrad (b) violet, 54.2 mrad (c) The resolving power is improved, with the minimum
resolvable angle becoming 60.0 mrad.
P38.14
Between 186 m and 271 m; blue is resolvable at larger distances because it has shorter
wavelength.
P38.16
30.5 m
P38.18
105 m
P38.20
514 nm
P38.22
1.81 µ m
P38.24
See the solution.
P38.26
74.2 grooves/mm
P38.28
(a) 0.738 mm
P38.30
93.4 pm
P38.32
5.51 m, 2.76 m, 1.84 m
(b) See the solution.
Diffraction Patterns and Polarization
403
P38.34
3
8
P38.36
1.11
P38.38
See the solution.
P38.40
See the solution.
P38.42
One slit 0.122 mm wide. The central maximum is twice as wide as the other maxima.
P38.44
See the solution.
P38.46
See the solution.
P38.48
5.63 mm. The assumption is absurd. Over a horizontal path of 12 km in air, density variations
associated with convection (“heat waves” or what an astronomer calls “seeing”) would make the
motorcycles completely unresolvable with any optical device.
P38.50
(a) 1.50 sec
P38.52
See the solution.
P38.54
(a) q = tan−1(n/1.33) − 36.9° (b) If we imagine n → ∞, then q → 53.1°. The material of the slab
in this case is higher in optical density than any gem. The light in the water skims along the upper
surface of the slab. (c) If we imagine n = 1, then q = 0. The slab is so low in optical density
that it is like air. The light strikes parallel surfaces as it enters and exits the water, both at the
polarizing angle.
P38.56
(a) See the solution.
P38.58
See the solution.
P38.60
(a) 1/8 (b) No recipes remain. The two experiments follow precisely analogous steps, but the
results are different. The middle filter in part (a) changes the polarization state of the light that
passes through it, but the recipe selection processes do not change individual recipes.
P38.62
(a) See the solution.
(b) 0.109°
P38.64
(a) See the solution.
(b) and (c) See the solution.
P38.66
(a) a sin θ = 1.430 3λ
(b) 0.189 ly
(c) 10.5 sec
(d) 1.52 mm
(b) 428 µ m
(b) a sin θ = 2.459 0 λ
39
Relativity
Note:
In chapters 39, 40, and 41 we use u to represent the speed of a particle with mass, reserving v for the
speeds associated with reference frames, wave functions, and photons.
ANSWERS TO QUESTIONS
CHAPTER OUTLINE
39.1
The Principle of Galilean
Relativity
39.2 The Michelson-Morley
Experiment
39.3 Einstein’s Principle of Relativity
39.4 Consequences of the Special
Theory of Relativity
39.5 The Lorentz Transformation
Equations
39.6 The Lorentz Velocity
Transformation Equations
39.7 Relativistic Linear Momentum
39.8 Relativistic Energy
39.9 Mass and Energy
39.10 The General Theory of Relativity
Q39.1
No. The principle of relativity implies that nothing can
travel faster than the speed of light in a vacuum, which
is 300 Mm/s. The electron would emit light in a conical
shock wave of Cerenkov radiation.
*Q39.2 Answer (c). The dimension parallel to the direction
of motion is reduced by the factor ␥ and the other
dimensions are unchanged.
*Q39.3
Answer (c). An oblate spheroid. The dimension in the
direction of motion would be measured to be scrunched in.
*Q39.4 Answer (e). The relativistic time dilation effect is
symmetric between the observers.
Q39.5
Suppose a railroad train is moving past you. One way to measure its length is this: You mark
the tracks at the cowcatcher forming the front of the moving engine at 9:00:00 AM, while your
assistant marks the tracks at the back of the caboose at the same time. Then you find the distance
between the marks on the tracks with a tape measure. You and your assistant must make the
marks simultaneously in your frame of reference, for otherwise the motion of the train would
make its length different from the distance between marks.
*Q39.6 (i) Answer (c). The Earth observer measures the clock in orbit to run slower.
(ii) Answer (b). They are not synchronized. They both tick at the same rate after return, but a
time difference has developed between the two clocks.
Q39.7
(a) Yours does.
(b) His does.
(c) If the velocity of relative motion is constant, both observers have equally valid views.
405
406
Chapter 39
Q39.8
Get a Mr. Tompkins book by George Gamow for a wonderful fictional exploration of this question.
Driving home in a hurry, you push on the gas pedal not to increase your speed by very much, but
rather to make the blocks get shorter. Big Doppler shifts in wave frequencies make red lights look
green as you approach them and make car horns and car radios useless. High-speed transportation
is very expensive, requiring huge fuel purchases. And it is dangerous, as a speeding car can knock
down a building. Having had breakfast at home, you return hungry for lunch, but you find you
have missed dinner. There is a five-day delay in transmission when you watch the Olympics in
Australia on live television. It takes ninety-five years for sunlight to reach Earth. We cannot see
the Milky Way; the fireball of the Big Bang surrounds us at the distance of Rigel or Deneb.
Q39.9
By a curved line. This can be seen in the middle of Speedo’s world-line in Figure 39.11, where he
turns around and begins his trip home.
Q39.10 A microwave pulse is reflected from a moving object. The waves that are reflected back are
Doppler shifted in frequency according to the speed of the target. The receiver in the radar gun
detects the reflected wave and compares its frequency to that of the emitted pulse. Using the
frequency shift, the speed can be calculated to high precision. Be forewarned: this technique
works if you are either traveling toward or away from your local law enforcement agent!
Q39.11
This system would be seen as a star moving in an elliptical path. Just like the light from a star in
a binary star system, the spectrum of light from the star would undergo a series of Doppler shifts
depending on the star’s speed and direction of motion relative to the observer. The repetition
rate of the Doppler shift pattern is the period of the orbit. Information about the orbit size can be
calculated from the size of the Doppler shifts.
Q39.12 According to p = γ mu , doubling the speed u will make the momentum of an object increase by
12
⎡ c2 − u 2 ⎤
the factor 2 ⎢ 2
.
2 ⎥
⎣ c − 4u ⎦
*Q39.13 From E2 ⫽ p2c2 ⫹ m2c4 ⫽ (mc2 ⫹ K )2 we consider pc = 2 Kmc 2 + K 2 . For a photon, pc ⫽ E.
For particles with mass, the greater the mass the greater the momentum if K is always 1 MeV.
The ranking is d > b > c > a.
Q39.14 As the object approaches the speed of light, its kinetic energy grows without limit. It would take
an infinite investment of work to accelerate the object to the speed of light.
*Q39.15 (i) Answer (a).
(ii) (c) and (iii) (d). There is no upper limit on the momentum or energy of an electron. As more
energy E is fed into the object without limit, its speed approaches the speed of light and its
momentum approaches E .
c
*Q39.16 Answer (b). Quasar light moves at three hundred million meters per second, just like the light
from a firefly at rest.
Q39.17 Any physical theory must agree with experimental measurements within some domain.
Newtonian mechanics agrees with experiment for objects moving slowly compared to the speed
of light. Relativistic mechanics agrees with experiment for objects at all speeds. Thus the two
theories must and do agree with each other for ordinary nonrelativistic objects. Both statements
given in the question are formally correct, but the first is clumsily phrased. It seems to suggest
that relativistic mechanics applies only to fast-moving objects.
Relativity
407
Q39.18 The point of intersection moves to the right. To state the problem precisely, let us assume that
each of the two cards moves toward the other parallel to the long dimension of the picture, with
2u
velocity of magnitude u. The point of intersection moves to the right at speed
= 2u cot φ ,
tan φ
where f is the small angle between the cards. As f approaches zero, cot φ approaches infinity.
Thus the point of intersection can move with a speed faster than c if v is sufficiently large and f
sufficiently small. For example, take u = 500 m s and φ = 0.000 19° . If you are worried about
holding the cards steady enough to be sure of the angle, cut the edge of one card along a curve so
that the angle will necessarily be sufficiently small at some place along the edge.
Let us assume the spinning flashlight is at the center of a grain elevator, forming a circular
screen of radius R. The linear speed of the spot on the screen is given by v = ω R, where w is the
angular speed of rotation of the flashlight. With sufficiently large w and R, the speed of the spot
moving on the screen can exceed c.
Neither of these examples violates the principle of relativity. Both cases are describing a point
of intersection: in the first case, the intersection of two cards and in the second case, the intersection of a light beam with a screen. A point of intersection is not made of matter so it has no
mass, and hence no energy. A bug momentarily at the intersection point could yelp, take a bite out
of one card, or reflect the light. None of these actions would result in communication reaching
another bug so soon as the intersection point reaches him. The second bug would have to wait for
sound or light to travel across the distance between the first bug and himself, to get the message.
As a child, the author used an Erector set to build a superluminal speed generator using the
intersecting-cards method. Can you get a visible dot to run across a computer screen faster than
light? Want’a see it again?
Q39.19 Special relativity describes inertial reference frames: that is, reference frames that are not
accelerating. General relativity describes all reference frames.
Q39.20 The downstairs clock runs more slowly because it is closer to the Earth and hence in a stronger
gravitational field than the upstairs clock.
SOLUTIONS TO PROBLEMS
Section 39.1
P39.1
The Principle of Galilean Relativity
The first observer watches some object accelerate under applied forces. Call the instantaneous
velocity of the object v1 . The second observer has constant velocity v 21 relative to the first, and
measures the object to have velocity v 2 = v1 − v 21.
dv
dv
The second observer measures an acceleration of
a2 = 2 = 1 .
dt
dt
This is the same as that measured by the first observer. In this nonrelativistic
case, they measure
the same forces as well. Thus, the second observer also confirms that ΣF = ma .
P39.2
The laboratory observer notes Newton’s second law to hold: F1 = ma1
(where the subscript 1 denotes the measurement was made in the laboratory frame of reference).
The observer in the accelerating frame measures the acceleration of the mass as a 2 = a1 − a ′
(where the subscript 2 implies the measurement was made in the accelerating frame of reference,
and the primed acceleration term is the acceleration of the accelerated frame with respect to the
laboratory frame of reference). If Newton’s second law held
accelerating frame, that
for the
observer would then find valid the relation F2 = ma 2 or F1 = ma 2
(since F1 = F2 and the mass is unchanged in each). But, instead, the accelerating frame observer
will find that F2 = ma 2 − ma ′, which is not Newton’s second law.
408
P39.3
Chapter 39
In the rest frame,
pi = m1 v1i + m2 v2i = ( 2 000 kg ) ( 20.0 m s ) + (1 500 kg ) ( 0 m s ) = 4.00 × 10 4 kg ⋅ m s
p f = ( m1 + m2 ) v f = ( 2 000 kg + 1 500 kg ) v f
Since pi = p f ,
vf =
4.00 × 10 4 kg ⋅ m s
= 11.429 m s
2 000 kg + 1 500 kg
In the moving frame, these velocities are all reduced by +10.0 m/s.
v1′i = v1i − v ′ = 20.0 m s − ( +10.0 m s ) = 10.0 m s
v2′i = v2i − v ′ = 0 m s − ( +10.0 m s ) = −10.0 m s
v ′f = 11.429 m s − ( +10.0 m s ) = 1.429 m s
Our initial momentum is then
pi′ = m1 v1′i + m2 v2′i = ( 2 000 kg ) (10.0 m s ) + (1 500 kg ) ( −10.0 m s ) = 5 000 kg ⋅ m s
and our final momentum has the same value:
p ′f = ( 2 000 kg + 1 500 kg ) v ′f = ( 3 500 kg ) (1.429 m s ) = 5 000 kg ⋅ m s
Section 39.2
The Michelson-Morley Experiment
Section 39.3
Einstein’s Principle of Relativity
Section 39.4
Consequences of the Special Theory of Relativity
P39.4
L = Lp
⎛ L⎞
v = c 1− ⎜ ⎟
⎝ Lp ⎠
v2
1− 2
c
2
2
⎛ Lp 2⎞
1
= c 1 − = 0.866c
Taking L =
where L p = 1.00 m gives v = c 1 − ⎜
⎟
2
4
⎝ Lp ⎠
Lp
P39.5
P39.6
∆t =
12
∆t p
so
⎡1 − ( v c )2 ⎤
⎣
⎦
12
For
∆t = 2 ∆t p
(a)
γ =
⎡ ⎛ ∆t p ⎞ 2 ⎤
v = c ⎢1 − ⎜
⎟ ⎥
⎢⎣ ⎝ ∆t ⎠ ⎥⎦
12
⎡ ⎛ ∆t ⎞ 2 ⎤
p
v = c ⎢1 − ⎜
⎟ ⎥
∆
2
t
⎢⎣ ⎝
p⎠ ⎥
⎦
1
1 − ( v c)
2
=
1
1 − ( 0.500 )
2
=
12
1
= c ⎡⎢1 − ⎤⎥
⎣ 4⎦
= 0.866c
2
3
The time interval between pulses as measured by the Earth observer is
∆t = γ∆t p =
2 ⎛ 60.0 s ⎞
= 0.924 s
3 ⎝ 75.0 ⎠
Thus, the Earth observer records a pulse rate of
(b)
60.0 s min
= 64.9 min .
0.924 s
At a relative speed v = 0.990 c, the relativistic factor g increases to 7.09 and the pulse rate
recorded by the Earth observer decreases to 10.6 min . That is, the life span of the
astronaut (reckoned by the duration of the total number of his heartbeats) is much longer as
measured by an Earth clock than by a clock aboard the space vehicle.
Relativity
P39.7
∆t p
∆t = γ∆t p =
1 − v2 c2
⎛
⎛
v2 ⎞
v2 ⎞
∆t p = ⎜ 1 − 2 ⎟ ∆t ≅ ⎜ 1 − 2 ⎟ ∆t
⎝
c ⎠
2c ⎠
⎝
so
⎛ v2 ⎞
∆t − ∆t p = ⎜ 2 ⎟ ∆t
⎝ 2c ⎠
1.00 × 10 6 m
v = 1 000 km h =
= 277.8 m s
3 600 s
v
= 9.26 × 10 −7
c
and
If
then
( ∆t − ∆t ) = ( 4.28 × 10 ) (3 600 s) = 1.54 × 10
−13
and
P39.8
For
(a)
p
v
= 0.990 , γ = 7.09
c
The muon’s lifetime as measured in the Earth’s rest frame is
∆t =
4.60 km
0.990c
and the lifetime measured in the muon’s rest frame is
∆t p =
(b)
P39.9
409
⎤
1 ⎡
4.60 × 10 3 m
∆t
=
⎢
⎥ = 2.18 µs
8
γ
7.09 ⎢⎣ 0.990 ( 3.00 × 10 m s ) ⎥⎦
2
L p 4.60 × 10 3 m
v
L = Lp 1 − ⎛ ⎞ =
=
= 649 m
⎝ c⎠
γ
7.09
The spaceship is measured by the Earth observer to be length-contracted to
L = Lp 1 −
v2
c2
or
⎛
v2 ⎞
L2 = L2p ⎜ 1 − 2 ⎟
⎝
c ⎠
Also, the contracted length is related to the time required to pass overhead by:
v2
( ct )2
c2
2
v2
2 v
Equating these two expressions gives L2p − L2p 2 = ( ct ) 2
c
c
2
v
⎡⎣ L2p + ( ct )2 ⎤⎦ 2 = L2p
or
c
L = vt
or
L2 = v 2 t 2 =
Using the given values:
L p = 300 m
this becomes
(1.41 × 10
giving
v = 0.800 c
5
and
m2 )
t = 7.50 × 10 −7 s
v2
= 9.00 × 10 4 m 2
c2
−9
s = 1.54 ns
410
P39.10
Chapter 39
(a)
The spaceship is measured by Earth observers to be of length L, where
L = Lp 1 −
v2
c2
v∆t = L p 1 −
v2
c2
and
L = v∆t
and
⎛
v2 ⎞
v 2 ∆t 2 = L2p ⎜ 1 − 2 ⎟
⎝
c ⎠
⎛
L2p ⎞
Solving for v, v 2 ⎜ ∆t 2 + 2 ⎟ = L2p
c ⎠
⎝
v=
cL p
c ∆t 2 + L2p
2
(b)
The tanks move nonrelativistically, so we have v =
(c)
For the data in problem 9,
c ( 300 m )
v=
(3 × 10
8
m s ) ( 0.75 × 10
2
−6
s ) + ( 300 m )
2
2
=
300 m
= 4.00 m s .
75 s
c ( 300 m )
2252 + 300 2 m
= 0.800 c
in agreement with problem 9. For the data in part (b),
c ( 300 m )
c ( 300 m )
v=
=
2
(3 × 108 m s ) ( 75 s )2 + (300 m )2 ( 2.25 × 1010 )2 + 300 2 m
= 1.33 × 10 −8 c = 4.00 m s
in agreement with part (b).
P39.11
We find Cooper’s speed:
GMm m v 2
=
r2
r
Solving,
⎡ GM ⎤
v=⎢
⎥
⎣ ( R + h) ⎦
12
Then the time period of one orbit is
(a)
T=
⎡ ( 6.67 × 10 −11 ) ( 5.98 × 10 24 ) ⎤
=⎢
6
6 ⎥
⎢⎣ ( 6.337 × 10 + 0.160 × 10 ) ⎥⎦
12
= 7.82 km s
6
2π ( R + h ) 2π ( 6.53 × 10 )
=
= 5.25 × 10 3 s
v
7.82 × 10 3
⎡⎛
v2 ⎞
The time difference for 22 orbits is ∆t − ∆t p = (γ − 1) ∆t p = ⎢⎜ 1 − 2 ⎟
c ⎠
⎢⎣⎝
−1 2
⎤
− 1⎥ ( 22T )
⎥⎦
2
⎛
1 v2
1 ⎛ 7.82 × 10 3 m s ⎞
⎞
)
∆t − ∆t p ≈ ⎜ 1 +
−
1
22
T
=
22 ( 5.25 × 10 3 s ) = 39.2 µs
(
⎟⎠
⎝
2 c2
2 ⎜⎝ 3 × 108 m s ⎟⎠
(b)
For each one orbit Cooper aged less by ∆t − ∆t p =
accurate to one digit .
P39.12
γ =
1
1 − ( v2 c2 )
= 1.01
so
v = 0.140 c
39.2 µs
= 1.78 µs. The press report is
22
Relativity
P39.13
(a)
Since your ship is identical to his, and you are at rest with respect to your own ship, its
length is 20.0 m .
P39.14
(b)
His ship is in motion relative to you, so you measure its length contracted to 19.0 m .
(c)
We have
L = Lp 1 −
from which
L 19.0 m
v2
=
= 0.950 = 1 − 2 and v = 0.312c
L p 20.0 m
c
v2
c2
In the Earth frame, Speedo’s trip lasts for a time
∆t =
20.0 ly
∆x
=
= 21.05 yr
v
0.950 ly yr
Speedo’s age advances only by the proper time interval
∆t p =
∆t
= 21.05 yr 1 − 0.952 = 6.574 yr during his trip
γ
Similarly for Goslo,
∆t p =
v2
20.0 ly
∆x
1− 2 =
1 − 0.752 = 17.64 yr
v
0.750 ly yr
c
While Speedo has landed on Planet X and is waiting for his brother, he ages by
20.0 ly
20.0 ly
−
= 5.614 yr
0.750 ly yr 0.950 ly yr
Then Goslo ends up older by 17.64 yr − ( 6.574 yr + 5.614 yr ) = 5.45 yr .
P39.15
The orbital speed of the Earth is as described by ΣF = ma:
v=
GmS
=
r
(6.67 × 10
−11
N ⋅ m 2 kg2 ) (1.99 × 10 30 kg )
1.496 × 1011 m
GmS mE mE v 2
=
r2
r
= 2.98 × 10 4 m s
The maximum frequency received by the extraterrestrials is
fobs = fsource
1 + ( 2.98 × 10 4 m s ) ( 3.00 × 108 m s )
1+ v c
= ( 57.0 × 10 6 Hz )
1− v c
1 − ( 2.98 × 10 4 m s ) ( 3.00 × 108 m s )
= 57.005 66 × 10 6 Hz
The minimum frequency received is
fobs = fsource
1 − ( 2.98 × 10 4 m s ) ( 3.00 × 108 m s )
1− v c
= ( 57.0 × 10 6 Hz )
1+ v c
1 + ( 2.98 × 10 4 m s ) ( 3.00 × 108 m s )
= 56.994 34 × 10 6 Hz
The difference, which lets them figure out the speed of our planet, is
( 57.005 66 − 56.994 34 ) × 106
Hz = 1.13 × 10 4 Hz
411
412
P39.16
Chapter 39
(a)
Let fc be the frequency as seen by the car. Thus,
fc = fsource
and, if f is the frequency of the reflected wave,
f = fc
(b)
(c)
(c + v)
(c − v)
Using the above result,
f ( c − v ) = fsource ( c + v )
which gives
( f − fsource ) c = ( f + fsource ) v ≈ 2 fsource v
The beat frequency is then
fbeat = f − fsource =
fbeat =
λ=
P39.17
c+v
c−v
f = fsource
Combining gives
c+v
c−v
( 2 ) ( 30.0 m s ) (10.0 × 10 9 Hz )
3.00 × 10 m s
8
c
=
fsource
fbeat λ
2
=
( 2 ) ( 30.0 m s )
( 0.030 0 m )
2 fsource v
2v
=
λ
c
= 2 000 Hz = 2.00 kHz
3.00 × 108 m s
= 3.00 cm
10.0 × 10 9 Hz
∆v =
∆fbeat λ ( 5 Hz ) ( 0.030 0 m )
=
= 0.075 0 m s ≈ 0.2 mi h
2
2
(d)
v=
(a)
When the source moves away from an observer, the observed frequency is
so
⎛ c − vs ⎞
fobs = fsource ⎜
⎝ c + vs ⎟⎠
12
where vs = vsource
When vs << c, the binomial expansion gives
⎛ c − vs ⎞
⎜⎝ c + v ⎟⎠
s
12
12
v ⎤
⎡
= ⎢1 − ⎛ s ⎞ ⎥
⎣ ⎝ c ⎠⎦
⎡ ⎛ vs ⎞ ⎤
⎢⎣1 + ⎝ c ⎠ ⎥⎦
−1 2
v
v
v
≈ ⎛1 − s ⎞ ⎛1 − s ⎞ ≈ ⎛1 − s ⎞
⎝
2c ⎠ ⎝
2c ⎠ ⎝
c⎠
v
fobs ≈ fsource ⎛ 1 − s ⎞
⎝
c⎠
So,
The observed wavelength is found from c = λobs fobs = λ fsource :
λobs =
λ fsource
λ fsource
λ
≈
=
fobs
fsource (1 − vs c ) 1 − vs c
⎛ v c ⎞
⎛ 1
⎞
∆λ = λobs − λ = λ ⎜
−1 = λ ⎜ s
⎝ 1 − vs c ⎟⎠
⎝ 1 − vs c ⎟⎠
Since 1 −
(b)
vs
≈ 1,
c
∆λ vsource
≈
λ
c
20.0 nm ⎞
⎛ ∆λ ⎞
vsource = c ⎜
= c⎛
= 0.050 4 c
⎝ 397 nm ⎠
⎝ λ ⎟⎠
Relativity
P39.18
413
For the light as observed
c
1+ v c
1+ v c c
fobs =
=
fsource =
1− v c
1 − v c λsource
λobs
1 + v c λsource 650 nm
=
=
1− v c
520 nm
λobs
1+
1+ v c
= 1.252 = 1.562
1− v c
v
v
= 1.562 − 1.562
c
c
v 0.562
=
= 0.220
c 2.562
v = 0.220 c = 6.59 × 10 7 m s
Section 39.5
P39.19
The Lorentz Transformation Equations
Let Suzanne be fixed in reference from S and see the two light-emission events with coordinates
x1 = 0, t1 = 0, x2 = 0, t 2 = 3 µs. Let Mark be fixed in reference frame S′ and give the events
coordinate x1′ = 0 , t1′ = 0 , t 2′ = 9 µs.
(a)
Then we have
v
⎛
t2′ = γ t2 − 2 x 2 ⎞
⎝
⎠
c
1
9 µs =
1 − v2 c
( 3 µs − 0 )
2
v2 8
=
c2 9
(b)
P39.20
γ =
1−
v2 1
=
c2 3
v = 0.943c
⎛ 3 × 108 m s ⎞
3
x2′ = γ ( x2 − vt 2 ) = 3 ( 0 − 0.943c × 3 × 10 −6 s ) ⎜
⎟⎠ = 2.55 × 10 m
⎝
c
1
1− v c
2
2
=
1
1 − 0.9952
= 10.0
We are also given: L1 = 2.00 m, and θ = 30.0° (both measured
in a reference frame moving relative to the rod).
Thus,
L1x = L1 cos θ1 = ( 2.00 m ) ( 0.867 ) = 1.73 m
and
L1y = L1 sin θ1 = ( 2.00 m ) ( 0.500 ) = 1.00 m
L2 x
γ
FIG. P39.20
L2 x is a proper length, related to L1x by
L1x =
Therefore,
L2 x = 10.0 L1x = 17.3 m
and
L2 y = L1y = 1.00 m
(Lengths perpendicular to the motion are unchanged.)
( L2 x ) 2 + ( L2 y )
(a)
L2 =
(b)
θ 2 = tan −1
L2 y
L2 x
2
gives
L2 = 17.4 m
gives
θ 2 = 3.30°
414
P39.21
Chapter 39
Einstein’s reasoning about lightning striking the ends of a train shows that the moving observer
sees the event toward which she is moving, event B, as occurring first. The S-frame coordinates
of the events we may take as ( x = 0, y = 0, z = 0, t = 0) and ( x = 100 m, y = 0, z = 0, t = 0).
Then the coordinates in S ′ are given by the Lorentz transformation. Event A is at ( x ′ = 0 ,
y ′ = 0, z ′ = 0, t ′ = 0). The time of event B is
v
1
80 m ⎞
⎛ 0 − 0.8c (100 m )⎞ = 1.667 ⎛ −
−7
t ′ = γ ⎛ t − 2 x⎞ =
2
⎜⎝ 3 × 108 m s ⎟⎠ = −4.44 × 10 s
2 ⎝
⎝
⎠
c ⎠
c
1 − 0.8
The time elapsing before A occurs is 444 ns .
P39.22
(a)
From the Lorentz transformation, the separations between the blue-light and red-light
events are described by
0 = γ ⎡⎣ 2.00 m − v (8.00 × 10 −9 s ) ⎤⎦
∆x ′ = γ ( ∆x − v∆t )
v=
(b)
2.00 m
= 2.50 × 108 m s
8.00 × 10 −9 s
γ =
1
1 − ( 2.50 × 10 m s )
8
2
(3.00 × 10
8
m s)
2
= 1.81
Again from the Lorentz transformation,
x ′ = γ ( x − vt ):
x ′ = 1.81 ⎡⎣3.00 m − ( 2.50 × 108 m s ) (1.00 × 10 −9 s ) ⎦⎤
x ′ = 4.97 m
(c)
v
t ′ = γ ⎛ t − 2 x⎞ :
⎝
c ⎠
⎡
( 2.50 × 108 m s ) (3.00 m )⎤⎥
t ′ = 1.81 ⎢1.00 × 10 −9 s −
⎢⎣
⎥⎦
(3.00 × 108 m s )2
t ′ = −1.33 × 10 −8 s
Section 39.6
P39.23
u x′ =
The Lorentz Velocity Transformation Equations
ux − v
−0.750 c − 0.750 c
= −0.960 c
=
1 − u x v c 2 1 − ( −0.750 ) ( 0.750 )
speed = 0.960 c
FIG. P39.23
P39.24
u x = Enterprise velocity
v = Klingon velocity
From Equation 39.16
u x′ =
0.900 c − 0.800 c
ux − v
=
= 0.357c
2
1 − ux v c
1 − ( 0.900 ) ( 0.800 )
FIG. P39.24
Relativity
Section 39.7
P39.25
(a)
Relativistic Linear Momentum
p = γ mu ; for an electron moving at 0.010 0c,
γ =
1
1 − (u c )
2
=
1
1 − ( 0.010 0 )
2
= 1.000 05 ≈ 1.00
p = 1.00 ( 9.11 × 10 −31 kg ) ( 0.010 0 ) ( 3.00 × 108 m s )
Thus,
p = 2.73 × 10 −24 kg ⋅ m s
(b)
(c)
P39.26
Following the same steps as used in part (a),
we find at 0.500c, γ = 1.15 and
p = 1.58 × 10 −22 kg ⋅ m s
At 0.900c, γ = 2.29 and
p = 5.64 × 10 −22 kg ⋅ m s
= γ mu
p=
we find the difference ∆ from the classical momentum, mu:
∆p = γ mu − mu = (γ − 1) mu
The difference is 1.00% when (γ − 1) mu = 0.010 0γ mu :
γ =
2
(b)
1 − (u c )
u = 0.141c
The difference is 10.0% when (γ − 1) mu = 0.100γ mu :
γ =
2
p − mu γ mu − mu
=
= γ − 1:
mu
mu
2
1
1
=
2
0.990
1 − (u c )
u
2
thus 1 − ⎛ ⎞ = ( 0.990 ) , and
⎝ c⎠
u
2
thus 1 − ⎛ ⎞ = ( 0.900 ) and
⎝ c⎠
P39.27
mu
Using the relativistic form,
(a)
415
1
1
=
2
0.900
1 − (u c )
u = 0.436c
γ −1 =
2
1
1 − (u c )
2
1 u
1 u
−1 ≈ 1+ ⎛ ⎞ −1 = ⎛ ⎞
2 ⎝ c⎠
2 ⎝ c⎠
2
p − mu 1 ⎛ 90.0 m s ⎞
= ⎜
= 4.50 × 10 −14
mu
2 ⎝ 3.00 × 108 m s ⎟⎠
2
416
Chapter 39
*P39.28 We can express the proportionality of the speeding fine to the excess momentum as F = d ( p − p )
where F is the fine, d is a proportionality constant, p is the magnitude of the vehicle’s momentum,
and p is the magnitude of its momentum as it travels at 90 km/h.
(a)
METHOD ONE. For low speeds, the classical expression for momentum is accurate. Our
equation describes the two cases
$80 = d ⎡⎣ m (190 km h ) − m ( 90 km h ) ⎤⎦ and Fa = d ⎡⎣ m (1 090 km h ) − m ( 90 km h ) ⎤⎦
Dividing gives
m (1 000 km h )
Fa
⇒ Fa = $800 .
=
m (100 km h )
$80
METHOD TWO. The relativistic momentum expression is always accurate:
⎛
⎞
m (190 km h )
m ( 90 km h )
$80 = d ⎜
−
⎟
2
⎜⎝ 1 − (190 km h )2 c 2
1 − ( 90 km h ) c 2 ⎟⎠
⎛
⎞
m (1 090 km h )
m (190 km h )
Fa = d ⎜
−
⎟
2
⎜⎝ 1 − (1 090 km h )2 c 2
1 − (190 km h ) c 2 ⎟⎠
To three or even to six digits, the answer is the same: Fa = $80 (10 ) = $800 .
(b)
Now the high-speed case must be described relativistically. The speed of light is
3 × 108 m s = 1.08 × 10 9 km h .
$80 = dm (100 km h )
⎛
⎞
m (1.000 000 09 × 10 9 km h )
⎟
Fb = d ⎜
90
m
km
h
−
(
)
⎜ 1 − (1.000 000 09 × 10 9 1.08 × 10 9 )2
⎟
⎝
⎠
Fb = dm ( 2.65 × 10 9 km h )
again dividing,
Fb 2.65 ×109
=
⇒ Fb = $2.12 ×109 .
2
$80
10
Relativity
P39.29
Relativistic momentum of the system of fragments must be conserved. For total momentum to be
zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and
subscript 1 to the lighter, p2 = p1
or
γ 2 m2 u2 = γ 1 m1u1 =
or
(1.67 × 10
−27
2.50 × 10 −28 kg
1 − ( 0.893)
kg ) u2
1 − ( u2 c )
2
× ( 0.893c )
= ( 4.960 × 10 −28 kg ) c
2
2
⎛ 1.67 × 10 −27 u2 ⎞
u22
=
−
1
⎜⎝ 4.960 × 10 −28 c ⎟⎠
c2
Proceeding to solve, we find
12.3
Section 39.8
*P39.30 (a)
u22
= 1 and u2 = 0.285c
c2
Relativistic Energy
K = E − E R = 5E R
E = 6 ER = 6 ( 9.11 × 10 −31 kg ) ( 3.00 × 108 m s ) = 4.92 × 10 −13 J = 3.07 MeV
2
(b)
E = γ mc 2 = γ ER
Thus γ =
P39.31
417
E
1
=6=
which yields
ER
1 − u 2 c2
⎛
1
ΣW = K f − K i = ⎜
⎜
⎝ 1− vf c
(
⎛
1
or ΣW = ⎜
⎜
⎝ 1− vf c
(
(a)
)
2
−
)
2
u = 0.986c
⎞
⎛
⎞
1
⎟ mc 2
− 1⎟ mc 2 − ⎜
⎟
⎜⎝ 1 − ( vi c )2 ⎟⎠
⎠
⎞
⎟ mc 2
2 ⎟
1 − ( vi c ) ⎠
1
⎛
⎞
2
1
1
−27
8
ΣW = ⎜
−
⎟ (1.673 × 10 kg ) ( 2.998 × 10 m s )
2
2
1 − ( 0.500 ) ⎠
⎝ 1 − ( 0.750 )
ΣW = 5.37 × 10 −11 J
(b)
⎛
⎞
2
1
1
−27
8
ΣW = ⎜
−
⎟ (1.673 × 10 kg ) ( 2.998 × 10 m s )
2
2
1 − ( 0.500 ) ⎠
⎝ 1 − ( 0.995)
ΣW = 1.33 × 10 −9 J
418
P39.32
Chapter 39
⎛
⎞
1
The relativistic kinetic energy of an object of mass m and speed u is K r = ⎜
− 1⎟ mc 2 .
2
2
⎝ 1− u c
⎠
⎛
⎞
1
Kr = ⎜
− 1⎟ mc 2 = 0.005 038 mc 2
−
.
1
0
010
0
⎝
⎠
For u = 0.100 c ,
The classical equation K c =
1
mu 2 gives
2
Kc =
1
2
m ( 0.100 c ) = 0.005 000 mc 2
2
0.005 038 − 0.005 000
= 0.751%
0.005 038
different by
For still smaller speeds the agreement will be still better.
P39.33
E = γ mc 2 = 2 mc 2 or
γ =2
2
⎛ 1⎞
3
u
Thus, = 1 − ⎜ ⎟ =
or
2
c
⎝γ ⎠
u=
The momentum is then
⎛ c 3 ⎞ ⎛ mc 2 ⎞
p = γ mu = 2 m ⎜
=
3
⎝ 2 ⎟⎠ ⎜⎝ c ⎟⎠
c 3
2
p=⎛
⎝
*P39.34 (a)
Using the classical equation,
MeV
938.3 MeV ⎞
3 = 1.63 × 10 3
⎠
c
c
K=
2
1 2 1
mu = ( 78.0 kg ) (1.06 × 10 5 m s )
2
2
= 4.38 × 1011 J
(b)
Using the relativistic equation,
⎛
⎞
1
1
K =⎜
−
⎟ mc 2
⎟⎠
⎜⎝ 1 − ( u c )2
⎡
⎤
2
1
⎢
1
K=
− ⎥ ( 78.0 kg ) ( 2.998 × 108 m s ) = 4.38 × 1011 J
⎢
⎥
5
8 2
⎣ 1 − (1.06 × 10 2.998 × 10 )
⎦
(c)
−1 2
u
When << 1, the binomial series expansion gives
c
⎡ ⎛ u⎞2 ⎤
⎢1 − ⎜⎝ ⎟⎠ ⎥
c ⎦
⎣
Thus,
⎡ ⎛ u⎞2 ⎤
⎢1 − ⎜⎝ ⎟⎠ ⎥
c ⎦
⎣
1 ⎛ u⎞
≈ 1+ ⎜ ⎟
2 ⎝ c⎠
−1 2
−1≈
2
1 ⎛ u⎞
⎜ ⎟
2 ⎝ c⎠
2
2
1 ⎛ u⎞
1
and the relativistic expression for kinetic energy becomes K ≈ ⎜ ⎟ mc 2 = mu 2 .
2 ⎝ c⎠
2
That is, in the limit of speeds much smaller than the speed of light, the relativistic and
classical expressions yield the same results. In this situation the two kinetic energy values
are experimentally indistinguishable. The fastest-moving macroscopic objects launched
by human beings move sufficiently slowly compared to light that relativistic corrections to
their energy are negligible.
Relativity
P39.35
P39.36
)
)
(a)
ER = mc 2 = (1.67 × 10 −27 kg ( 2.998 × 10 8 m s
(b)
E = γ mc 2 =
(c)
K = E − mc 2 = 4.81 × 10 −10 J − 1.50 × 10 −10 J = 3.31 × 10 −10 J = 2.07 × 10 3 MeV
1.50 × 10 −10 J
2
419
= 1.50 × 10 −10 J = 938 MeV
= 4.81 × 10 −10 J = 3.00 × 10 3 MeV
⎡1 − ( 0.950 c c )2 ⎤
⎣
⎦
12
We must conserve both energy and relativistic momentum of the
system of fragments. With subscript 1 referring to the 0.868 c
particle and subscript 2 to the 0.987 c particle,
γ1 =
1
1 − ( 0.868 )
2
= 2.01 and γ 2 =
1
1 − ( 0.987 )
= 6.22
2
Conservation of energy gives E1 + E2 = Etotal
which is
γ 1 m1c 2 + γ 2 m2 c 2 = mtotal c 2
or
2.01m1 + 6.22 m2 = 3.34 × 10 −27 kg
This reduces to:
m1 + 3.09 m2 = 1.66 × 10 −27 kg
(1)
FIG. P39.36
Since the final momentum of the system must equal zero, p1 = p2
gives
γ 1 m1u1 = γ 2 m2 u2
or
( 2.01) ( 0.868c ) m1 = ( 6.22 ) ( 0.987c ) m2
which becomes
m1 = 3.52 m2
(2)
Solving (1) and (2) simultaneously, 3.52 m2 + 3.09 m2 = 1.66 × 10 −27 kg
m1 = 8.84 × 10 −28 kg and m2 = 2.51 × 10 −28 kg
P39.37
p = γ mu
E = γ mc 2
E 2 = (γ mc
)
p 2 = (γ mu )
2 2
E − p c = (γ mc
2
2 2
) − (γ mu )
2 2
2
2
c =γ
2
2
(( mc ) − ( mc) u ) = ( mc )
2 2
2
2
2 2
⎛
u2 ⎞ ⎛
u2 ⎞
1
−
1
−
⎜⎝
c 2 ⎟⎠ ⎜⎝
c 2 ⎟⎠
as was to be demonstrated.
P39.38
(a)
q ( ∆V ) = K = (γ − 1) me c 2
Thus, γ =
1
1 − (u c )
2
= 1+
25 000 eV
q ( ∆V )
= 1+
= 1.0489
2
me c
511 000 eV
V
so 1 – (u/c)2 = 0.9089 and u = 0.302c
(b)
)
)
K = (γ − 1) me c 2 = q ( ∆V ) = (1.60 × 10 −19 C ( 2.50 × 10 4 J C = 4.00 × 10 −15 J
−1
= ( mc 2
)
2
420
Chapter 39
⎛
⎞
1
1
K
K + mc 2
*P39.39 From K = (γ − 1) mc 2 = ⎜
− 1⎟ mc 2 we have
+
1
=
=
.
2
2
mc 2
mc 2
1 − u 2 c2
⎝ 1− u c
⎠
1−
u2
m2 c4
=
c 2 ( K + mc 2
)
mc 2
(
u2
= 1−
c2
( K + mc2
)
2
2
)
2
⎛ ⎛ mc 2 ⎞ 2 ⎞
u = c ⎜1 − ⎜
2⎟ ⎟
⎝ ⎝ K + mc ⎠ ⎠
12
12
(a)
⎛ ⎛ 0.511⎞ 2 ⎞
Electron: u = c ⎜ 1 − ⎜
⎟ ⎟
⎝ ⎝ 2.511⎠ ⎠
(b)
⎛ ⎛ 938 ⎞ 2 ⎞
Proton: u = c ⎜ 1 − ⎜
⎟ ⎟
⎝ ⎝ 940 ⎠ ⎠
(c)
0.979 c − 0.065 2 c = 0.914 c = 2.74 × 10 8 m s
= 0.979 c
12
= 0.065 2 c
In this case the electron is moving relativistically, but the classical expression
1
mv 2 is
2
accurate to two digits for the proton.
(d)
⎛ ⎛ 0.511 ⎞ 2 ⎞
Electron: u = c ⎜ 1 − ⎜
⎟ ⎟
⎝ ⎝ 2 000.511⎠ ⎠
⎛ ⎛ 938 ⎞ 2 ⎞
Proton: u = c ⎜ 1 − ⎜
⎟ ⎟
⎝ ⎝ 2 938 ⎠ ⎠
12
= 0.999 999 97 c
12
= 0.948 c
Excess speed = 0.999 999 97 c − 0.948 c = 0.052 3c = 1.57 × 10 7 m s
As the kinetic energies of both particles become large, the difference in their speeds
approaches zero. By contrast, classically the speed difference would become large without
any finite limit.
P39.40
(a)
E = γ mc 2 = 20.0 GeV with mc 2 = 0.511 MeV for electrons.
20.0 × 10 9 eV
= 3.91 × 10 4 .
0.511 × 10 6 eV
1
γ =
= 3.91 × 10 4 from which u = 0.999 999 999 7 c
1 − (u/c)2
Thus, γ =
(b)
(c)
2
L p 3.00 × 10 3 m
⎛ u⎞
L = Lp 1 − ⎜ ⎟ =
=
= 7.67 × 10 −2 m = 7.67 cm
⎝ c⎠
γ
3.91 × 10 4
Relativity
P39.41
421
Conserving total momentum of the decaying particle system, pbefore decay = pafter decay = 0
pv = pµ = γ mµ u = γ ( 207 me ) u
Conservation of mass-energy for the system gives Eµ + Ev = Eπ : γ mµ c 2 + pv c = mπ c 2
pv
= 273me
c
u
γ ( 207 me ) + γ ( 207 me ) = 273me
c
γ ( 207 me ) +
Substituting from the momentum equation above,
or
Then,
u
273
γ ⎛1 + ⎞ =
= 1.32:
⎝
⎠
c
207
1+ u c
= 1.74
1− u c
)
K µ = (γ − 1) mµ c 2 = (γ − 1) 207 ( me c 2 :
u
= 0.270
c
⎛
⎞
1
Kµ = ⎜
− 1⎟ 207 ( 0.511 MeV)
⎜⎝ 1 − ( 0.270 )2
⎟⎠
K µ = 4.08 MeV
Also,
Ev = mπ c 2 − γ mµ c 2 = ( 273 − 207γ ) me c 2
Ev = Eπ − Eµ:
⎛
⎞
207
Ev = ⎜ 273 −
⎟ ( 0.511 MeV )
2
1 − ( 0.270 ) ⎠
⎝
Ev = 29.6 MeV
(
*P39.42 K = (γ − 1) mc 2 = (1 − u 2 c 2
)
−1/ 2
)
− 1 mc 2 . We use the series expansion from Appendix B.5:
K = mc 2 ⎡⎣1 + ( − 12 ) (−u 2 / c 2 ) + ( − 12 ) ( − 32 ) 21! (−u 2 / c 2 )2 + − 1⎤⎦
K = 12 mu 2 + 83 m
u4
+ The actual kinetic energy, given by this relativistic equation, is greater
c2
than the classical (1/2)mu2. The difference, for m = 1 000 kg and u = 25 m/s, is
3
8
(1 000 kg)
Section 39.9
(25 m/s) 4
= 1.6 × 10 −9 J
( 3 × 10 8 m/s)2
~ 10 −9 J
Mass and Energy
*P39.43 E = 2.86 × 10 5 J leaves the system so the final mass is smaller . The mass-energy relation says
E
2.86 × 10 5 J
−12
that E = mc 2 . Therefore,
m= 2 =
kg
2 = 3.18 × 10
8
c
×
3.00
10
m
s
(
)
A mass loss of this magnitude, as a fraction of a total of 9.00 g, could not be detected .
P39.44
∆m =
7
9
E P ∆t 0.800 (1.00 × 10 J s ) ( 3.00 yr ) ( 3.16 × 10 s yr )
=
=
= 0.842 kg
c2
c2
(3.00 × 108 m s )2
422
P39.45
Chapter 39
P=
2
dE d ( mc )
dm
=
= c2
= 3.85 × 10 26 W
dt
dt
dt
Thus,
P39.46
)
2 me c 2 = 1.02 MeV
Section 39.10
P39.47
dm
3.85 × 10 26 J s
= 4.28 × 10 9 kg s
=
dt ( 3.00 × 10 8 m s 2
(a)
Eγ ≥ 1.02 MeV
The General Theory of Relativity
For the satellite ΣF = ma:
GM E m m v 2 m ⎛ 2π r ⎞
=
=
r2
r
r ⎝ T ⎠
2
GM E T 2 = 4π 2 r 3
)
⎛ 6.67 × 10 −11 N ⋅ m 2 ( 5.98 × 10 24 kg ( 43 080 s )2 ⎞
r=⎜
⎟
kg 2 4π 2
⎝
⎠
13
r = 2.66 × 10 7 m
7
2π r 2π ( 2.66 × 10 m )
=
= 3.87 × 10 3 m s
T
43 080 s
(b)
v=
(c)
The small fractional decrease in frequency received is equal in magnitude to the fractional
increase in period of the moving oscillator due to time dilation:
⎛
1
fractional change in f = − (γ − 1) = − ⎜
⎜
3
8
⎝ 1 − ( 3.87 × 10 3 × 10
)
2
⎞
− 1⎟
⎟
⎠
2
⎛
1 ⎡ ⎛ 3.87 × 10 3 ⎞ ⎤⎞
= 1 − ⎜1 − ⎢− ⎜
⎥⎟ = −8.34 × 10 −11
⎜⎝
2 ⎢⎣ ⎝ 3 × 10 8 ⎟⎠ ⎥⎦⎟⎠
(d)
The orbit altitude is large compared to the radius of the Earth, so we must use U g = −
∆U g = −
)
6.67 × 10 −11 Nm 2 ( 5.98 × 10 24 kg m
kg 2 2.66 × 10 7 m
+
= 4.76 × 10 7 J kg m
∆f ∆U g 4.76 × 10 7 m 2 s 2
=
=
= +5.29 × 10 −10
2
8
f
mc 2
3
×
10
m
s
(
)
(e)
−8.34 × 10 −11 + 5.29 × 10 −10 = +4.46 × 10 −10
)
6.67 × 10 −11 Nm ( 5.98 × 10 24 kg m
kg 6.37 × 10 6 m
GM E m
.
r
Relativity
Additional Problems
P39.48
(a)
dearth = vt earth = vγ tastro
1−
1=
so
v2 ⎛ v ⎞
=
(1.50 × 10 −5 )
c2 ⎝ c ⎠
v2
(1 + 2.25 × 10 −10 )
c2
( 2.00 × 10
1
1 − v2 c2
30.0 yr
−10
2
v 2 v ( 2.25 × 10 )
=
2
2
c
c
1−
so
yr ) c = v
6
−1 2
v
1
= (1 + 2.25 × 10 −10 ) = 1 − ( 2.25 × 10 −10 )
c
2
v
= 1 − 1.12 × 10 −10
c
(b)
⎛
⎞ 2 ⎛ 2.00 × 10 6 yr ⎞
2
1
1
− 1⎟ (1 000 ) (1 000 kg ) ( 3 × 108 m s )
K =⎜
−
⎟ mc = ⎜
2
2
⎝
⎠
30
yr
⎠
⎝ 1− v c
= 6.00 × 10 27 J
P39.49
(c)
13¢ ⎞ ⎛ k ⎞ ⎛ W ⋅ s ⎞ ⎛ h ⎞
= $2.17 × 10 20
6.00 × 10 27 J = 6.00 × 10 27 J ⎛
⎝ kWh ⎠ ⎝ 10 3 ⎠ ⎝ J ⎠ ⎜⎝ 3 600 s ⎟⎠
(a)
1013 MeV = (γ − 1) m p c 2
so
vp ≈ c
P39.50
γ = 1010
t′ =
t 10 5 yr
=
= 10 −5 yr ~ 10 2 s
γ
1010
(b)
d ′ = ct ′ ~108 km
(a)
When K e = K p ,
me c 2 (γ e − 1) = m p c 2 γ p − 1
In this case,
me c 2 = 0.511 MeV, m p c 2 = 938 MeV
and
Substituting,
(b)
(
−1 2
2
γ e = ⎡⎣1 − ( 0.750 ) ⎤⎦ = 1.511 9
m c 2 (γ e − 1)
( 0.511 MeV ) (1.511 9 − 1)
γ p = 1+ e
= 1+
2
938 MeV
mpc
but
= 1.000 279
1
γp =
12
⎡1 − u c 2 ⎤
p
⎣
⎦
Therefore,
u p = c 1 − γ −p2 = 0.023 6 c
When pe = p p
γ p m p u p = γ e me u e
Thus,
γ pu p =
and
up
(
c
which yields
)
)
or γ p u p =
γ e me u e
mp
(1.511 9 ) ( 0.511 MeV c2 ) ( 0.750c )
= 6.177 2 × 10
938 MeV c 2
−4
⎛ up ⎞
1− ⎜ ⎟
⎝ c⎠
2
u p = 6.18 × 10 −4 c = 185 km s
= 6.177 2 × 10 −4 c
423
424
Chapter 39
*P39.51 (a)
We let H represent K /mc 2 . Then H + 1 =
1− u / c
2
(b)
(c)
u approaches c as K increases without limit.
(d)
a=
2
2
1 − u 2 /c 2 =
so
⎛ H 2 + 2H ⎞
and u = c ⎜ 2
⎝ H + 2 H + 1⎟⎠
1
u
H + 2H
= 1− 2
= 2
2
c
H + 2H + 1 H + 2H + 1
u goes to 0 as K goes to 0.
2
1
H + 2H + 1
2
1/ 2
du
dt
1 ⎛ H 2 + 2H ⎞
=c ⎜ 2
2 ⎝ H + 2 H + 1⎟⎠
−1/ 2
⎛ H 2 + 2 H + 1⎞
= a = c⎜
⎝ H 2 + 2 H ⎟⎠
P39.52
1
⎛ ⎡ H 2 + 2 H + 1⎤ [ 2 H + 2 ] − ⎡ H 2 + 2 H ⎤ [ 2 H + 2 ] ⎞ d ( K /mc 2 )
⎣
⎦
⎣
⎦
⎟
⎜
2
⎟⎠
⎜⎝
dt
⎡⎣ H 2 + 2 H + 1⎤⎦
1/ 2
⎛ H +1 ⎞ P
P
⎜⎝ [ H + 1]4 ⎟⎠ mc 2 = mcH 1/ 2 ( H + 2)1/ 2 ( H + 1)2
P
P
(e)
=
When H is small we have approximately a =
1/ 2 , in agreement with the
mcH 1/ 2 21/ 2 ( 2 mK )
nonrelativistic case.
(f)
When H is large the acceleration approaches P / mcH 3 = P m2c5/K 3.
(g)
As energy is steadily imparted to particle, the particle’s acceleration decreases. It decreases
steeply, proportionally to 1/K 3 at high energy. In this way the particle’s speed cannot reach
or surpass a certain upper limit, which is the speed of light in vacuum.
(a)
Since Mary is in the same reference frame, S ′, as Ted, she measures the ball to have the
same speed Ted observes, namely u x′ = 0.800 c .
Lp
(b)
∆t ′ =
(c)
L = Lp 1 −
u x′
=
1.80 × 1012 m
= 7.50 × 10 3 s
0.800 ( 3.00 × 108 m s )
v2
( 0.600 c )
= (1.80 × 1012 m ) 1 −
= 1.44 × 1012 m
c2
c2
2
Since v = 0.600 c and u x′ = − 0.800 c, the velocity Jim measures for the ball is
ux =
(d)
Jim measures the ball and Mary to be initially separated by 1.44 × 1012 m. Mary’s motion
at 0.600c and the ball’s motion at 0.385c nibble into this distance from both ends. The gap
closes at the rate 0.600 c + 0.385 c = 0.985 c, so the ball and catcher meet after a time
∆t =
P39.53
( − 0.800c ) + ( 0.600c )
u x′ + v
=
= − 0.385c
2
1 + u x′ v c
1 + ( − 0.800 ) ( 0.600 )
1.44 × 1012 m
= 4.88 × 10 3 s
0.985 ( 3.00 × 108 m s )
∆mc 2 4 ( 938.78 MeV ) − 3 728.4 MeV
=
× 100% = 0.712%
mc 2
4 ( 938.78 MeV )
Relativity
P39.54
The energy of the first fragment is given by E12 = p12 c 2 + ( m1c 2
)
2
= (1.75 MeV) + (1.00 MeV)
2
425
2
E1 = 2.02 MeV
For the second, E22 = ( 2.00 MeV) + (1.50 MeV)
2
(a)
2
E2 = 2.50 MeV
Energy is conserved, so the unstable object had E = 4.52 MeV. Each component of
momentum is conserved, so the original object moved
2
2
⎛ 1.75 MeV ⎞ ⎛ 2.00 MeV ⎞
with p 2 = px2 + py2 = ⎜
⎟⎠ . Then for
⎟⎠ + ⎜⎝
⎝
c
c
it ( 4.52 MeV) = (1.75 MeV) + ( 2.00 MeV) + ( mc 2
2
(b)
P39.55
2
2
Now E = γ mc 2 gives 4.52 MeV =
1
1 − v2 c2
)
2
m=
3.65 MeV 1 −
3.65 MeV
c2
v2
= 0.654 , v = 0.589c
c2
Look at the situation from the instructors’ viewpoint since they are at rest relative to the clock,
and hence measure the proper time. The Earth moves with velocity v = −0.280 c relative to the
instructors while the students move with a velocity u ′ = −0.600 c relative to Earth. Using the
velocity addition equation, the velocity of the students relative to the instructors (and hence the
clock) is:
u=
(a)
v + u′
( −0.280 c ) − ( 0.600 c )
=
= −0.753c (students relative to clock)
1 + vu ′ c 2 1 + ( −0.280 c ) ( −0.600 c ) c 2
With a proper time interval of ∆t p = 50.0 min, the time interval measured by the students is:
∆t = γ ∆ t p
with
γ =
1
1 − ( 0.753c ) c 2
2
= 1.52
Thus, the students measure the exam to last T = 1.52 ( 50.0 min ) = 76.0 minutes .
(b)
The duration of the exam as measured by observers on Earth is:
∆t = γ ∆ t p with γ =
*P39.56 (a)
1
1 − ( 0.280 c ) c 2
2
so T = 1.04 ( 50.0 min ) = 52.1 minutes
The speed of light in water is c/1.33, so the electron’s speed is 1.1c/1.33. Then
γ =
1
(b)
= 1.779 and the energy is g mc2 = 1.779(0.511 MeV) = 0.909 MeV
1 − (1.11
/ .33)2
K = 0.909 MeV – 0.511 MeV = 0.398 MeV
(c)
pc = E 2 − ( mc 2 )2 = γ 2 − 1 mc 2 = 1.779 2 − 1 0.511 MeV = 0.752 MeV
p = 0.752 MeV/c = 4.01 × 10−22 kg ⋅ m/s
(d)
sin q = v/u where v is the wave speed, which is the speed of light in water, and u is the
source speed. Then sin q = 1/1.1 = 0.909 so q = 65.4°
426
P39.57
Chapter 39
(a)
Take the spaceship as the primed frame, moving toward the right at v = +0.600 c .
ux =
Then u x′ = +0.800 c, and
Lp
u x′ + v
0.800 c + 0.600 c
=
= 0.946c
2
1 + ( u x′ v ) c
1 + ( 0.800 ) ( 0.6600 )
L = ( 0.200 ly ) 1 − ( 0.600 ) = 0.160 ly
(b)
L=
(c)
The aliens observe the 0.160-ly distance closing because the probe nibbles into it from one
end at 0.800c and the Earth reduces it at the other end at 0.600c.
(d)
γ
2
:
0.160 ly
= 0.114 yr
0.800c + 0.600 c
Thus,
time =
⎛
⎞
1
K =⎜
− 1⎟ mc 2:
2
2
⎠
⎝ 1− u c
⎛
⎞
1
K =⎜
− 1⎟ ( 4.00 × 10 5 kg ( 3.00 × 10 8 m s
⎜⎝ 1 − ( 0.946 )2
⎟⎠
)
)
2
K = 7.50 × 10 22 J
*P39.58 (a)
From equation 39.18, the speed of light in the laboratory frame is
u=
(b)
v + c/n
c (1 + nv/c)
=
2
1 + vc/nc
n (1 + v/nc)
When v is much less than c we have
u=
v
c⎛
nv
1 + ⎞ ⎛1 + ⎞
⎝
⎠
⎝
n
c
nc ⎠
−1
≈
v
v
c⎛
nv
c
nv v ⎞ c
1 + ⎞ ⎛1 − ⎞ ≈ ⎛1 +
−
= +v− 2
⎠
⎝
⎠
⎝
⎠
⎝
n
c
nc
n
c nc
n
n
The Galilean velocity transformation equation 4.20 would indeed give c/n + v for the speed
of light in the moving water. The third term –v/n2 does represent a relativistic effect that
was observed decades before the Michelson-Morley experiment. It is a piece of twentiethcentury physics that dropped into the nineteenth century. We could say that light is
intrinsically relativistic.
(c)
To take the limit as v approaches c we must go back to
u=
P39.59
c (1 + nv/c)
n (1 + v/nc)
to find u approaches
c (1 + nc/c) c(1 + n)
= c
=
n (1 + c/nc)
n +1
The observer measures the proper length of the tunnel, 50.0 m, but measures the train contracted
to length
L = Lp 1 −
shorter than the tunnel by
v2
2
= 100 m 1 − ( 0.950 ) = 31.2 m
c2
50.0 − 31.2 = 18.8 m so it is completely within the tunnel .
Relativity
P39.60
427
If the energy required to remove a mass m from the surface is equal to its rest energy mc 2 ,
then
GM s m
= mc 2
Rg
and
Rg =
)
−11
N ⋅ m 2 kg 2 (1.99 × 10 30 kg
GM s ( 6.67 × 10
=
2
c2
( 3.00 × 108 m s
)
)
Rg = 1.47 × 10 3 m = 1.47 km
P39.61
(a)
At any speed, the momentum of the particle is given by
mu
p = γ mu =
Since F = qE =
dp
:
dt
qE =
1 − (u c )
−1 2
d ⎡ ⎛
u2 ⎞ ⎤
⎢ mu ⎜ 1 − 2 ⎟ ⎥
dt ⎢⎣ ⎝
c ⎠ ⎥⎦
⎛
u2 ⎞
qE = m ⎜ 1 − 2 ⎟
⎝
c ⎠
⎛
du 1
u2 ⎞
+ mu ⎜ 1 − 2 ⎟
⎝
dt 2
c ⎠
−3 2
⎛ 2u ⎞ du
⎝ c 2 ⎠ dt
)
du qE ⎛
u2 ⎞
1− 2 ⎟
a=
=
⎜
dt
m ⎝
c ⎠
and
32
qE
. As u
m
approaches c, the acceleration approaches zero, so that the object can never reach the speed
of light.
For u small compared to c, the relativistic expression reduces to the classical a =
u
(c)
−1 2
qE du ⎡ 1 − u 2 c 2 + u 2 c 2 ⎤
⎢
⎥
=
m
dt ⎢ (1 − u 2 c 2 3 2 ⎥
⎣
⎦
So
(b)
2
du
∫ (1 − u
0
t
2
c
t
)
2 32
=
t
x = ∫ udt = qEc ∫
0
qE
dt
m
t =0
∫
0
u=
tdt
m c +q E t
2 2
2
2 2
x=
qEct
m c + q2 E 2t 2
2 2
c
qE
(
m 2 c 2 + q 2 E 2 t 2 − mc
)
428
P39.62
Chapter 39
(a)
An observer at rest relative to the mirror sees the light travel a distance D = 2 d − x , where
x = vt S is the distance the ship moves toward the mirror in time t S . Since this observer
agrees that the speed of light is c, the time for it to travel distance D is
tS =
(b)
D 2d − vt S
=
c
c
tS =
2d
c+v
The observer in the rocket measures a length-contracted initial distance to the mirror of
L = d 1−
v2
c2
and the mirror moving toward the ship at speed v. Thus, he measures the distance the light
vt
travels as D = 2 ( L − y ) where y =
is the distance the mirror moves toward the ship
2
before the light reflects from it. This observer also measures the speed of light to be c, so
the time for it to travel distance D is:
t=
*P39.63 (a)
D 2⎡
v 2 vt ⎤
2d
2d
= ⎢ d 1 − 2 − ⎥ so ( c + v ) t =
( c + v ) ( c − v ) or t =
c c ⎢⎣
c
2 ⎥⎦
c
c
c−v
c+v
Conservation of momentum γ mu:
mu
1− u c
2
2
+
m ( −u )
3 1− u c
2
2
M vf
=
1− v c
2
f
2
=
2 mu
3 1 − u 2 c2
Conservation of energy γ mc 2:
mc 2
1 − u 2 c2
+
mc 2
3 1 − u 2 c2
Mc 2
=
1 − v 2f c 2
To start solving we can divide: v f =
M
1 − u 4c
2
M=
(b)
2
=
4m
3 1− u c
2
2
=
=
4 mc 2
3 1 − u 2 c2
2u u
= . Then
4
2
M
(1 2)
4 − u 2 c2
2m 4 − u 2 c2
3 1 − u 2 c2
4m
, in agreement with the classical result, which is the
3
arithmetic sum of the masses of the two colliding particles.
When u << c , this reduces to M =
Relativity
P39.64
429
Take the two colliding protons as the system
E1 = K + mc 2
E2 = mc 2
E12 = p12 c 2 + m 2 c 4
p2 = 0
In the final state, E f = K f + Mc 2 :
E 2f = p 2f c 2 + M 2 c 4
By energy conservation, E1 + E2 = E f , so
E12 + 2 E1 E2 + E22 = E 2f
)
p12 c 2 + m 2 c 4 + 2 ( K + mc 2 mc 2 + m 2 c 4
= p c +M c
2 2
f
2 4
By conservation of momentum,
p1 = p f
Then
M 2 c 4 = 2 Kmc 2 + 4 m 2 c 4
=
4 Km 2 c 4
+ 4 m 2 c4
2 mc 2
Mc 2 = 2 mc 2 1 +
K
2 mc 2
FIG. P39.64
By contrast, for colliding beams we have
E1 = K + mc 2
In the original state,
E2 = K + mc 2.
In the final state,
E f = Mc 2
E1 + E2 = E f :
K + mc 2 + K + mc 2 = Mc 2
K ⎞
⎛
Mc 2 = 2 mc 2 ⎜ 1 +
⎟
⎝
2 mc 2 ⎠
P39.65
We choose to write down the answer to part (b) first.
(b)
Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti,
stationary with respect to both. Just as our spaceship is passing him, he also sees the blast
waves from both explosions. Judging both stars to be stationary, this observer concludes
that the two stars blew up simultaneously .
(a)
We in the spaceship moving past the hermit do not calculate the explosions to be
simultaneous. We measure the distance we have traveled from the Sun as
2
v
2
L = L p 1 − ⎛ ⎞ = ( 6.00 ly ) 1 − ( 0.800 ) = 3.60 ly
⎝ c⎠
We see the Sun flying away from us at 0.800c while the light from the Sun approaches at
1.00c. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time
we calculate to have elapsed since the Sun exploded is
3.60 ly
= 2.00 yr
1.80c
continued on next page
430
Chapter 39
We see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only
0.200c faster. We measure the gap between that star and its blast wave as 3.60 ly and
growing at 0.200c. We calculate that it must have been opening for
3.60 ly
= 18.0 yr
0.200c
and conclude that Tau Ceti exploded 16.0 years before the Sun .
P39.66
Take m = 1.00 kg.
2
1
1
⎛ u⎞
⎛ u⎞
mu 2 = mc 2 ⎜ ⎟ = ( 4.50 × 1016 J ⎜ ⎟
⎝ c⎠
⎝ c⎠
2
2
)
2
The classical kinetic energy is
Kc =
and the actual kinetic energy is
⎛
⎛
⎞
⎞
1
1
Kr = ⎜
− 1⎟ mc 2 = ( 9.00 × 1016 J ⎜
− 1⎟
⎜⎝ 1 − ( u c )2
⎜⎝ 1 − ( u c )2
⎟⎠
⎟⎠
)
u
c
Kc ( J)
Kr ( J)
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
0.990
0.000
0.045 × 10166
0.180 × 1016
0.405 × 1016
0.720 × 1016
1.13 × 1016
1.62 × 1016
2.21 × 1016
2.88 × 1016
3.65 × 1016
4.411 × 1016
0.000
0.0453 × 1016
0.186 × 1016
0.435 × 1016
0.820 × 1016
1.39 × 1016
2.25 × 1016
3.60 × 1016
6.00 × 1016
11.6 × 1016
54.8 × 1016
K c = 0.990 K r , when
FIG. P39.66
2
⎡
⎤
1 ⎛ u⎞
1
− 1⎥, yielding u = 0.115 c .
⎜⎝ ⎟⎠ = 0.990 ⎢⎢
2
⎥
2 c
⎣ 1 − (u c )
⎦
Similarly,
K c = 0.950 K r
when
u = 0.257 c
and
K c = 0.500 K r
when
u = 0.786 c
Relativity
P39.67
Since the total momentum is zero before decay, it is necessary that after the decay
pnucleus = pphoton =
Eγ
c
=
14.0 keV
c
Also, for the recoiling nucleus, E 2 = p 2 c 2 + ( mc 2
Thus,
( mc
So
1+
2
+K
)
2
= (14.0 keV) + ( mc 2
2
)
2
)
2
with mc 2 = 8.60 × 10 −9 J = 53.8 GeV.
2
2
2
K
1 ⎛ 14.0 keV ⎞
⎛ 14.0 keV ⎞
≈ 1+ ⎜
= 1+ ⎜
⎟ (Binomial Theorem)
⎝ mc 2 ⎟⎠
mc 2
2 ⎝ mc 2 ⎠
K≈
and
2
K ⎞
⎛ 14.0 keV ⎞
⎛
+1
=⎜
or ⎜ 1 +
2⎟
⎝ mc 2 ⎟⎠
⎝
mc ⎠
(14.0 keV)2
2 mc 2
(14.0 × 10 eV) =
=
2 ( 53.8 × 10 eV)
2
3
9
1.82 × 10 −3 eV
ANSWERS TO EVEN PROBLEMS
P39.2
See the solution.
P39.4
0.866c
P39.6
(a) 64.9/min
P39.8
(a) 2.18 µs b) The muon sees the planet surface moving 649 m up toward it.
P39.10
(a)
(b) 10.6/min
cL p
(b) 4.00 m s
P39.12
c 2 ∆t 2 + L2p
v = 0.140 c
P39.14
5.45 yr, Goslo is older
P39.16
(c) 2.00 kHz
P39.18
0.220 c = 6.59 × 107 m/s
P39.20
(a) 17.4 m
P39.22
(a) 2.50 × 108 m/s
P39.24
0.357c
P39.26
(a) 0.141c
P39.28
(a) $800
P39.30
P39.32
(c) See the solution.
(d) ± 0.075 0 m/s ≈ 0.2 mi/h
(b) 3.30°
(b) 4.97 m
(b) 0.436 c
(b) $2.12 × 109
(a) 3.07 MeV (b) 0.986c
See the solution.
(c) –1.33 × 10−8 s
431
432
Chapter 39
P39.34
(a) 438 GJ (b) 438 GJ (c) The two kinetic energy values are experimentally indistinguishable,
essentially identical. The fastest-moving macroscopic objects launched by human beings move
sufficiently slowly compared to light that relativistic corrections to their energy are negligible.
P39.36
8.84 × 10–28 kg and 2.51 × 10–28 kg
P39.38
(a) 0.302 c (b) 4.00 fJ
P39.40
(a) 3.91 × 104
P39.42
larger by ~10–9 J
P39.44
0.842 kg
P39.46
1.02 MeV
P39.48
(a)
P39.50
(a) 0.023 6 c
P39.52
(a) 0.800 c
P39.54
(a)
P39.56
(b) u = 0.999 999 999 7 c
v
= 1 − 1.12 × 10 −10 (b) 6.00 × 10 27 J
c
(c) 7.67 cm
(c) $2.17 × 10 20
(b) 6.18 × 10 −4 c
(b) 7.50 ks
3.65 MeV
c2
(c) 1.44 Tm, –0.385 c
(d) 4.88 ks
(b) v = 0.589c
(a) 0.909 MeV (b) 0.398 MeV (c) 0.752 MeV/c = 4.01 × 10−22 kg ⋅ m/s
P39.58
(c) u → c
P39.60
1.47 km
P39.62
(a)
P39.64
See the solution.
P39.66
See the solution, 0.115 c, 0.257 c, 0.786 c.
2d
c+v
(b)
2d
c
c−v
c+v
(d) 65.4°
40
Introduction to Quantum Physics
In chapters 39, 40, and 41 we use u to represent the speed of a particle with mass, reserving v for the
speeds associated with reference frames, wave functions, and photons.
Note:
CHAPTER OUTLINE
40.1
40.2
40.3
40.4
40.5
40.6
40.7
40.8
Blackbody Radiation and
Planck’s Hypothesis
The Photoelectric Effect
The Compton Effect
Photons and Electromagnetic
Waves
The Wave Properties of Particles
The Quantum Particle
The Double-Slit Experiment
Revisited
The Uncertainty Principle
Q40.2
ANSWERS TO QUESTIONS
Q40.1
The first flaw is that the Rayleigh–Jeans law predicts
that the intensity of short wavelength radiation emitted
by a blackbody approaches infinity as the wavelength
decreases. This is known as the ultraviolet catastrophe.
The second flaw is the prediction of much more power
output from a blackbody than is shown experimentally.
The intensity of radiation from the blackbody is given by
the area under the red I ( λ , T ) vs. l curve in Figure 40.5
in the text, not by the area under the blue curve.
Planck’s Law dealt with both of these issues and
brought the theory into agreement with the experimental
data by adding an exponential term to the denominator
1
that depends on λ . This both keeps the predicted
intensity from approaching infinity as the wavelength
decreases and keeps the area under the curve finite.
Our eyes are not able to detect all frequencies of electromagnetic waves. For example, all
objects that are above 0 K in temperature emit electromagnetic radiation in the infrared region.
This describes everything in a dark room. We are only able to see objects that emit or reflect
electromagnetic radiation in the visible portion of the spectrum.
*Q40.3 (i) The power input to the filament has increased by 8 × 2 = 16 times. The filament radiates this
greater power according to Stefan’s law, so its absolute temperature is higher by the fourth root
of 16. It is two times higher. Answer (d).
(ii) Wien’s displacement law then says that the wavelength emitted most strongly is half as large:
answer (f).
Q40.4
No. The second metal may have a larger work function than the first, in which case the incident
photons may not have enough energy to eject photoelectrons.
Q40.5
Comparing Equation 40.9 with the slope-intercept form of the equation for a straight line,
y = mx + b, we see that the slope in Figure 40.11 in the text is Planck’s constant h and that the
y intercept is −φ , the negative of the work function. If a different metal were used, the slope
would remain the same but the work function would be different. Thus, data for different metals
appear as parallel lines on the graph.
433
434
Chapter 40
Q40.6
Wave theory predicts that the photoelectric effect should occur at any frequency, provided the
light intensity is high enough. However, as seen in the photoelectric experiments, the light must
have a sufficiently high frequency for the effect to occur.
Q40.7
The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each of
them has gotten its energy from a single photon. According to Planck’s E = hf , the photon energy
depends on the frequency of the light. The intensity controls only the number of photons reaching
a unit area in a unit time.
Q40.8
Ultraviolet light has shorter wavelength and higher photon energy than visible light.
*Q40.9 Answer (c). UV light has the highest frequency of the three, and hence each photon delivers
more energy to a skin cell. This explains why you can become sunburned on a cloudy day: clouds
block visible light and infrared, but not much ultraviolet. You usually do not become sunburned
through window glass, even though you can see the visible light from the Sun coming through the
window, because the glass absorbs much of the ultraviolet and reemits it as infrared.
Q40.10 The Compton effect describes the scattering of photons from electrons, while the photoelectric
effect predicts the ejection of electrons due to the absorption of photons by a material.
*Q40.11 Answer (a). The x-ray photon transfers some of its energy to the electron. Thus, its frequency
must decrease.
Q40.12 A few photons would only give a few dots of exposure, apparently randomly scattered.
*Q40.13 (i) a and c. Some people would say that electrons and protons possess mass and photons do not.
(ii) a and c
(iii) a, b, and c
(iv) a, b, and c
(v) b
(vi) a, b, and c
Q40.14 Light has both classical-wave and classical-particle characteristics. In single- and double-slit
experiments light behaves like a wave. In the photoelectric effect light behaves like a particle.
Light may be characterized as an electromagnetic wave with a particular wavelength or
frequency, yet at the same time light may be characterized as a stream of photons, each carrying
a discrete energy, hf. Since light displays both wave and particle characteristics, perhaps it would
be fair to call light a “wavicle.” It is customary to call a photon a quantum particle, different from
a classical particle.
Q40.15 An electron has both classical-wave and classical-particle characteristics. In single- and doubleslit diffraction and interference experiments, electrons behave like classical waves. An electron
has mass and charge. It carries kinetic energy and momentum in parcels of definite size, as
classical particles do. At the same time it has a particular wavelength and frequency. Since an
electron displays characteristics of both classical waves and classical particles, it is neither a
classical wave nor a classical particle. It is customary to call it a quantum particle, but another
invented term, such as “wavicle,” could serve equally well.
Q40.16 The discovery of electron diffraction by Davisson and Germer was a fundamental advance in our
understanding of the motion of material particles. Newton’s laws fail to properly describe the
motion of an object with small mass. It moves as a wave, not as a classical particle. Proceeding
from this recognition, the development of quantum mechanics made possible describing the
motion of electrons in atoms; understanding molecular structure and the behavior of matter at
the atomic scale, including electronics, photonics, and engineered materials; accounting for the
motion of nucleons in nuclei; and studying elementary particles.
Introduction to Quantum Physics
435
p2
= q∆V , which is the same for both particles, then we see that the
2m
⎛
h⎞
electron has the smaller momentum and therefore the longer wavelength ⎜ λ = ⎟ .
⎝
p⎠
*Q40.17 Answer (a). If we set
Q40.18 Any object of macroscopic size—including a grain of dust—has an undetectably small
wavelength and does not exhibit quantum behavior.
*Q40.19 The wavelength is described by l = h/p in all cases, so e, f, and g are all the same. For the
photons the momentum is given by p = E/c, so a is also the same, and d has wavelength ten times
larger. For the particles with mass, pc = (E 2 – m2c4)1/2 = ([K + mc2]2 – m2c4)1/2 = (K 2 + 2Kmc2)1/2.
Thus a particle with larger mass has more momentum for the same kinetic energy, and a shorter
wavelength. The ranking is then d > a = e = f = g > b > c.
Q40.20 The intensity of electron waves in some small region of space determines the probability that an
electron will be found in that region.
1
µ m, while the de Broglie wavelength of an
2
electron can be 4 orders of magnitude smaller. Would your collar size be measured more precisely
1
with an unruled meter stick or with one engraved with divisions down to
mm?
10
Q40.21 The wavelength of violet light is on the order of
*Q40.22 Answer (c). The proton has 1836 time more momentum, thus more momentum uncertainty, and
thus possibly less position uncertainty.
Q40.23 The spacing between repeating structures on the surface of the feathers or scales is on the order of
1/2 the wavelength of light. An optical microscope would not have the resolution to see such fine
detail, while an electron microscope can. The electrons can have much shorter wavelength.
Q40.24 (a)
The slot is blacker than any black material or pigment. Any radiation going in through
the hole will be absorbed by the walls or the contents of the box, perhaps after several
reflections. Essentially none of that energy will come out through the hole again. Figure 40.1
in the text shows this effect if you imagine the beam getting weaker at each reflection.
(b)
The open slots between the glowing tubes are brightest. When you look into a slot, you
receive direct radiation emitted by the wall on the far side of a cavity enclosed by the
fixture; and you also receive radiation that was emitted by other sections of the cavity wall
and has bounced around a few or many times before escaping through the slot. In Figure
40.1 in the text, reverse all of the arrowheads and imagine the beam getting stronger at
each reflection. Then the figure shows the extra efficiency of a cavity radiator. Here is the
conclusion of Kirchhoff’s thermodynamic argument: . . . energy radiated. A poor reflector—a
good absorber—avoids rising in temperature by being an efficient emitter. Its emissivity
is equal to its absorptivity: e = a . The slot in the box in part (a) of the question is a blackbody with reflectivity zero and absorptivity 1, so it must also be the most efficient possible
radiator, to avoid rising in temperature above its surroundings in thermal equilibrium. Its
emissivity in Stefan’s law is 100% = 1, higher than perhaps 0.9 for black paper, 0.1 for lightcolored paint, or 0.04 for shiny metal. Only in this way can the material objects underneath
these different surfaces maintain equal temperatures after they come to thermal equilibrium
and continue to exchange energy by electromagnetic radiation. By considering one blackbody facing another, Kirchhoff proved logically that the material forming the walls of the
cavity made no difference to the radiation. By thinking about inserting color filters between
two cavity radiators, he proved that the spectral distribution of blackbody radiation must be
a universal function of wavelength, the same for all materials and depending only on the
temperature. Blackbody radiation is a fundamental connection between the matter and the
energy that physicists had previously studied separately.
436
Chapter 40
SOLUTIONS TO PROBLEMS
Section 40.1
Blackbody Radiation and Planck’s Hypothesis
2.898 × 10 −3 m ⋅ K
= 5.18 × 10 3 K
560 × 10 −9 m
P40.1
T=
*P40.2
(a)
λmax =
(b)
The wavelength emitted most strongly is infrared (greater than 700 nm), and much more
energy is radiated at wavelengths longer than lmax than at shorter wavelengths.
*P40.3
2.898 × 10 −3 m ⋅ K
= 999 nm
2900 K
The peak radiation occurs at approximately 560 nm wavelength. From Wien’s displacement law,
T=
0.289 8 × 10 −2 m ⋅ K 0.289 8 × 10 −2 m ⋅ K
=
≈ 5 200 K
λmax
560 × 10 −9 m
Clearly, a firefly is not at this temperature, so this is not blackbody radiation .
P40.4
P40.5
(a)
λmax =
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
~
~ 10 −7 m
10 4 K
T
(b)
λmax ~
2.898 × 10 −3 m ⋅ K
~ 10 −10 m
10 7 K
ultraviolet
γ -ray
Planck’s radiation law gives intensity-per-wavelength. Taking E to be the photon energy and n
to be the number of photons emitted each second, we multiply by area and wavelength range to
have energy-per-time leaving the hole:
2π hc 2 ( λ2 − λ1 ) π ( d 2 )
= En = nhf
=
5 ⎛ 2 hc ⎡( λ + λ )kBT ⎤
⎞
⎡⎣( λ1 + λ2 ) 2 ⎤⎦ ⎜ e ⎣ 1 2 ⎦ − 1⎟
⎠
⎝
2
P
n=
P
E
=
where
E = hf =
2hc
λ1 + λ2
8π 2 cd 2 ( λ2 − λ1 )
2 hc ⎡ λ + λ k T ⎤
( λ1 + λ2 )4 ⎛⎜⎝ e ⎣( 1 2 ) B ⎦ − 1⎞⎟⎠
8π 2 ( 3.00 × 108 m s ) ( 5.00 × 10 −5 m ) (1.00 × 10 −9 m )
2
=
n=
(1 001 × 10
−9
⎡
−34
J⋅s )(3.00 ×108
4 ⎛ ⎢ 2(6.626 ×10
m ) ⎜ e⎣
⎝
5.90 × 1016 s
= 1.30 × 1015 s
( e3.84 − 1)
) (
)(
)(
)
m s ⎤⎥ ⎡⎢ 1 001×10−9 m 1.38 ×10−23 J K 7.50 ×103 K ⎤⎥
⎦ ⎣
⎦
⎞
− 1⎟
⎠
Introduction to Quantum Physics
P40.6
437
(a)
P = eAσ T 4 = 1( 20.0 × 10 −4 m 2 ) ( 5.67 × 10 −8 W m 2 ⋅ K 4 ) ( 5 000 K )4 = 7.09 × 10 4 W
(b)
λmaxT = λmax ( 5 000 K ) = 2.898 × 10 −3 m ⋅ K ⇒ λmax = 580 nm
(c)
We compute:
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
= 2.88 × 10 −6 m
=
kBT
(1.38 × 10 −23 J K ) (5 000 K )
The power per wavelength interval is
P ( λ ) = AI ( λ ) =
2π hc 2 A
, and
λ ⎡⎣ exp ( hc λ kBT ) − 1⎤⎦
5
2π hc 2 A = 2π ( 6.626 × 10 −34 ) ( 3.00 × 108 ) ( 20.0 × 10 −4 ) = 7.50 × 10 −19 J ⋅ m 4 s
2
P ( 580 nm ) =
(580 × 10
7.50 × 10 −19 J ⋅ m 4 s
−9
m ) ⎡⎣ exp ( 2.88 µ m 0.580 µ m ) − 1⎤⎦
5
=
1.15 × 1013 J m ⋅ s
e4.973 − 1
= 7.99 × 1010 W m
(d)–(i)
The other values are computed similarly:
hc
λ k BT
λ
(d)
(e)
(f )
(c)
(g)
(h)
(i)
( j)
ehc /λ kBT − 1
2π hc 2 A
λ5
P (λ ), W m
2882.6
7.96 × 101251 7.50 × 10 26 9.42 × 10 −1226
576.5
2.40 × 10 250 2.40 × 10 23 1.00 × 10 −227
7.21
1347
7.32 × 1013 5.44 × 1010
4.97
143.5
1.15 × 1013 7.99 × 1010
4.12
60.4
4.46 × 1012 7.38 × 1010
0.00288
0.00289
7.50 × 10 −4 0.260
2.88 × 10 −5 2.88 × 10 −5
7.50 × 10 −14 2.60 × 10 −99
1.00 nm
5.00 nm
400 nm
580 nm
700 nm
1.00 mm
100.0 cm
We approximate the area under the
as two trapezoids:
P ( λ ) versus l curve, between 400 nm and 700 nm,
⎡( 5.44 + 7.99 ) × 1010 W m ⎤⎦ ⎡⎣( 580 − 400 ) × 10 −9 m⎤⎤⎦
P=⎣
2
⎡( 7.99 + 7.38 ) × 10
+⎣
10
P = 2.13 × 10 4 W
P40.7
(a)
W m ⎤⎦ ⎡⎣( 700 − 580 ) × 10 −9 m ⎤⎦
2
so the power radiated as visible light is approximately 20 kW .
P = eAσ T 4 , so
14
⎛ P ⎞
T=
⎝ eAσ ⎠
(b)
λmax =
14
⎡
⎤
3.85 × 10 26 W
⎢
⎥
=
⎢ 1 ⎡ 4π ( 6.96 × 108 m )2 ⎤ ( 5.67 × 10 −8 W m 2 ⋅ K 4 ) ⎥
⎦
⎣ ⎣
⎦
= 5.78 × 10 3 K
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
=
= 5.01 × 10 −7 m = 501 nm
T
5.78 × 10 3 K
438
P40.8
Chapter 40
Energy of a single 500-nm photon:
Eγ = hf =
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
= 3.98 × 10 −19 J
−9
λ
500
×
10
m
(
)
The energy entering the eye each second
2
π
E = P∆t = IA∆t = ( 4.00 × 10 −11 W m 2 ) ⎡⎢ (8.50 × 10 −3 m ) ⎤⎥ (1.00 s ) = 2.27 × 10 −15 J
⎣4
⎦
The number of photons required to yield this energy
n=
P40.9
P40.10
2.27 × 10 −15 J
E
=
= 5.71 × 10 3 photons
Eγ 3.98 × 10 −19 J photon
(a)
1.00 eV ⎞
E = hf = ( 6.626 × 10 −34 J ⋅ s ) ( 620 × 1012 s −1 ) ⎛
= 2.57 eV
⎝ 1.60 × 10 −19 J ⎠
(b)
1.00 eV ⎞
E = hf = ( 6.626 × 10 −34 J ⋅ s ) ( 3.10 × 10 9 s −1 ) ⎛
= 1.28 × 10 −5 eV
⎝ 1.60 × 10 −19 J ⎠
(c)
1.00 eV ⎞
E = hf = ( 6.626 × 10 −34 J ⋅ s ) ( 46.0 × 10 6 s −1 ) ⎛
= 1.91 × 10 −7 eV
⎝ 1.60 × 10 −19 J ⎠
(d)
λ=
c 3.00 × 108 m s
=
= 4.84 × 10 −7 m = 484 nm, visible light ( blue )
f
620 × 1012 Hz
λ=
c 3.00 × 108 m s
=
= 9.68 × 10 −2 m = 9.68 cm, radio wave
f
3.10 × 10 9 Hz
λ=
c 3.00 × 108 m s
=
= 6.52 m, radio wave
46.0 × 10 6 Hz
f
We take θ = 0.030 0 radians. Then the pendulum’s total energy is
E = mgh = mg ( L − L cos θ )
E = (1.00 kg ) ( 9.80 m s 2 ) (1.00 − 0.999 5) = 4.41 × 10 −3 J
ω
1
=
2π 2π
g
= 0.498 Hz
L
The frequency of oscillation is
f =
The energy is quantized,
E = nhf
Therefore,
E
4.41 × 10 −3 J
=
n=
hf ( 6.626 × 10 −34 J ⋅ s ) ( 0.498 s −1 )
FIG. P40.10
= 1.34 × 10 31
P40.11
Each photon has an energy
E = hf = ( 6.626 × 10 −34 ) ( 99.7 × 10 6 ) = 6.61 × 10 −26 J
This implies that there are
150 × 10 3 J s
= 2.27 × 10 30 photons s
6.61 × 10 −26 J photon
Introduction to Quantum Physics
*P40.12 (a)
V=
439
4 3 4
π r = π ( 0.02 m )3 = 3.35 × 10 −5 m 3
3
3
m = ρ V = 7.86 × 10 3 kg m 3 3.35 × 10 −5 m 3 = 0.263 kg
(b)
A = 4π r 2 = 4π ( 0.02 m ) = 5.03 × 10 −3 m 2
2
P = σ AeT 4 = ( 5.67 × 10 −8 W m 2 K 4 ) 5.03 × 10 −3 m 2 0.86 ( 293 K )4 = 1.81 W
(c)
It emits but does not absorb radiation, so its temperature must drop according to
(
Q = mc∆T = mc T f − Ti
dT f
dt
(d)
=
)
dT f
dQ
= mc
dt
dt
−1.81 J s
dQ /dt −P
=
= −0.015 3 °C s
=
mc 0.263 kg 448 J kg ⋅ C°
mc
= −0.919 °C min
λmaxT = 2.898 × 10 −3 m ⋅ K
λmax =
2.898 × 10 −3 m ⋅ K
= 9.89 × 10 −6 m infrared
293 K
hc 6.63 × 10 −34 J ⋅ s 3 × 108 m s
=
= 2.01 × 10 −20 J
λ
9.89 × 10 −6 m
(e)
E = hf =
(f )
The energy output each second is carried by photons according to
P = ⎛⎝ ⎞⎠ E
∆t
N P
N
1.81 J s
= 8.98 × 1019 photon s
∆t E 2.01 × 10 −20 J photon
Matter is coupled to radiation quite strongly, in terms of photon numbers.
=
P40.13
Planck’s radiation law is
=
I (λ, T ) =
2π hc 2
λ 5 ( ehc λ kBT − 1
)
2
Using the series expansion
ex = 1 + x +
Planck’s law reduces to
I (λ, T ) =
3
x
x
+ +
2 ! 3!
2π ckBT
2π hc 2
2π hc 2
=
≈ 5
λ4
λ ⎡⎣(1 + hc λ kBT + ) − 1⎤⎦ λ ( hc λ kBT )
5
which is the Rayleigh-Jeans law, for very long wavelengths.
440
Chapter 40
Section 40.2
P40.14
(a)
(b)
The Photoelectric Effect
λc =
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
= 296 nm
=
φ
V ) (1.60 × 10 −19 J eV )
( 4.20 eV
fc =
c 3.00 × 108 m s
=
= 1.01 × 1015 Hz
λc
296 × 10 −9 m
hc
= φ + e∆VS :
λ
(6.626 × 10 ) (3.00 × 10 ) = ( 4.20 eV ) 1.60 × 10
(
180 × 10
−34
8
−19
−9
J eV )
+ (1.60 × 10 −19 ) ∆VS
∆VS = 2.71 V
Therefore,
P40.15
P40.16
(a)
e∆VS =
1 240 nm ⋅ eV
hc
−φ →φ =
− 0.376 eV = 1.990 eV
λ
546.1 nm
(b)
e∆VS =
1 240 nm ⋅ eV
hc
−φ =
− 1.90 eV → ∆VS = 0.216 V
λ
587.5 nm
K max =
2
1
1
2
mumax
= ( 9.11 × 10 −31 ) ( 4.60 × 10 5 ) = 9.64 × 10 −20 J = 0.602 eV
2
2
(a)
φ = E − K max =
(b)
fc =
*P40.17 (a)
1 240 eV ⋅ nm
− 0.602 nm = 1.38 eV
625 nm
⎛ 1.60 × 10 −19 J ⎞
φ
1.38 eV
14
=
⎟⎠ = 3.34 × 10 Hz
1 eV
V
h 6.626 × 10 −34 J ⋅ s ⎜⎝
We could find the energy of a photon with wavelength 400 nm and check whether it
hc
exceeds the work function. But we can instead use λc =
to find the threshold
φ
wavelength for each sample:
Li:
λc =
Be:
λc =
Hg:
λc =
J ⋅ s ) ( 3.00 × 108 m s )
( 2.30 eV ) (1.60 × 10 −19 J eV )
= 540 nm
J ⋅ s ) ( 3.00 × 108 m s )
( 3.90 eV ) (1.60 × 10 −19 J eV )
= 319 nm
J ⋅ s ) ( 3.00 × 108 m s )
( 4.50 eV ) (1.60 × 10 −19 J eV )
= 276 nm
(6.626 × 10
−34
(6.626 × 10
−34
(6.626 × 10
−34
We must have λ < λc for photocurrent to exist.
Thus, only lithium will exhibit the photoeleectric effect.
(b)
For lithium,
hc
= φ + K max
λ
(6.626 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s )
400 × 10 −9 m
K max = 1.29 × 10 −19 J = 0.806 eV
= ( 2.30 eV ) (1.60 × 10 −19 ) + K max
Introduction to Quantum Physics
P40.18
The energy needed is
E = 1.00 eV = 1.60 × 10 −19 J
The energy absorbed in time interval ∆t is
E = P∆t = IA∆t
441
1.60 × 10 −19 J
E
= 1.28 × 10 7 s = 148 days
=
IA ( 500 J s ⋅ m 2 ) ⎡π ( 2.82 × 10 −15 m )2 ⎤
⎣
⎦
The success of quantum mechanics contrasts with the gross failure of the classical theory of the
photoelectric effect.
so
P40.19
∆t =
Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximum kinetic
energy K max = hf − φ ,
or
K max =
(6.626 × 10
J ⋅ s ) ( 3.00 × 108 m s ) ⎛ 1.00 eV ⎞
− 4.70 eV = 1.51 eV
⎝ 1.60 × 10 −19 J ⎠
200 × 10 −9 m
−34
The sphere is left with positive charge and so with positive potential relative to V = 0 at r = ∞.
As its potential approaches 1.51 V, no further electrons will be able to escape, but will fall back
onto the sphere. Its charge is then given by
V=
P40.20
keQ
r
or
Q=
−2
rV ( 5.00 × 10 m ) (1.51 N ⋅ m C )
=
= 8.41 × 10 −12 C
8.99 × 10 9 N ⋅ m 2 C2
ke
(a)
By having the photon source move toward the metal, the incident photons are Doppler
shifted to higher frequencies, and hence, higher energy.
(b)
If v = 0.280 c,
f′= f
Therefore,
φ = ( 6.626 × 10 −34 J ⋅ s ) ( 9.33 × 1014 Hz ) = 6.18 × 10 −19 J = 3.87 eV
At v = 0.900 c,
f = 3.05 × 1015 Hz
(c)
1+ v c
1.28
= ( 7.00 × 1014 )
= 9.33 × 1014 Hz
1− v c
0.720
1.00 eV ⎞
and K max = hf − φ = ( 6.626 × 10 −34 J ⋅ s ) ( 3.05 × 1015 Hz ) ⎛
− 3.87 eV
⎝ 1.60 × 10 −19 J ⎠
= 8.78 eV
Section 40.3
P40.21
The Compton Effect
E=
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
= 2.84 × 10 −19 J = 1.78 eV
λ
700 × 10 −9 m
p=
h 6.626 × 10 −34 J ⋅ s
=
= 9.47 × 10 −28 kgg ⋅ m s
λ
700 × 10 −9 m
442
Chapter 40
h
(1 − cos θ ) we calculate the wavelength of the scattered photon.
me c
For example, at q = 30° we have
*P40.22 (a) and (b) From ∆λ =
λ ′ + ∆λ = 120 × 10 −12 +
6.626 × 10 −34
(1 − cos 30.0° ) = 120.3 × 10 −12 m
(9.11 × 10 −31 ) (3.00 × 108 )
The electron carries off the energy the photon loses:
Ke =
1 ⎞
hc hc 6.626 × 10 −34 J ⋅ s 3 × 108 m ⎛ 1
= 27.9 eV
−
=
−
−19
−12
⎝
λ0 λ ′ (1.6 × 10 J/eV ) s 10 m 120 120.3 ⎠
The other entries are computed similarly.
(c)
P40.23
p, degrees
0
30
60
90
120
150
180
k⬘, pm
120.0
120.3
121.2
122.4
123.6
124.5
124.8
Ke, eV
0
27.9
104
205
305
376
402
180°. We could answer like this: The photon imparts the greatest momentum to the
originally stationary electron in a head-on collision. Here the photon recoils straight
back and the electron has maximum kinetic energy.
E ′ = E0 − E ′
With K e = E ′, K e = E0 − E ′ gives
E′ =
E0
hc
and λ ′ =
2
E′
λ′ =
λ ′ = λ0 + λC (1 − cos θ )
hc
hc
=2
= 2λ0
E0 2
E0
2λ0 = λ0 + λC (1 − cos θ )
1 − cos θ =
P40.24
λ0 0.001 60
=
λC 0.002 43
θ = 70.0°
(a)
∆λ =
h
6.626 × 10 −34
(1 − cos θ ) : ∆λ =
( 1 − cos 37.0° ) = 4.88 × 10 −13 m
me c
(9.11 × 10 −31 ) (3.00 × 108 )
(b)
E0 =
hc
:
λ0
(300 × 10
3
eV ) (1.60 × 10 −19 J eV ) =
(6.626 × 10 ) (3.00 × 10
−34
λ0
λ0 = 4.14 × 10 −12 m
and
λ ′ = λ0 + ∆λ = 4.63 × 10 −12 m
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
E′ =
= 4.30 × 10 −14 J = 268 keV
=
λ′
4.63 × 10 −12 m
(c)
K e = E0 − E ′ = 300 keV − 268.5 keV = 31.5 keV
8
m s)
Introduction to Quantum Physics
P40.25
(a)
Conservation of momentum in the x direction gives:
pγ = pγ′ cos θ + pe cos φ
or since θ = φ ,
h ⎛
h
= ⎜ p + ⎞⎟ cos θ
λ0 ⎝ e λ ′ ⎠
Conservation of momentum in the y direction gives:
0 = pγ′ sin θ − pe sin θ
which (neglecting the trivial solution θ = 0) gives:
pe = pγ′ =
Substituting [2] into [1] gives:
h 2h
=
cos θ , or
λ0 λ ′
Then the Compton equation is
giving
Since Eγ =
hc
, this may be written as:
λ0
(b)
λ ′ = 2λ0 cos θ
me c 2 + Eγ
2 me c 2 + Eγ
=
Using Equation (3): Eγ′ =
h
(1 − cos θ )
me c
h
2λ0 cos θ − λ0 =
(1 − cos θ )
me c
λ ′ − λ0 =
hc 1
(1 − cos θ )
λ 0 me c 2
⎛ Eγ ⎞
2 cos θ − 1 = ⎜
(1 − cos θ )
⎝ m c 2 ⎟⎠
e
0.511 MeV + 0.880 MeV
= 0.732 so that θ = φ = 43.0°
1.022 MeV + 0.880 MeV
Eγ
0.880 MeV
hc
hc
=
=
=
λ ′ λ0( 2 cos θ ) 2 cos θ 2 cos 43.0°
= 0.602 MeV = 602 keV
(c)
Eγ′
0.602 MeV
= 3.21 × 10 −22 kg ⋅ m s
c
c
0.602 MeV
= 3.21 × 10 −22 kg ⋅ m s .
From Equation (2), pe = pγ′ =
c
Then,
[3]
Eγ ⎞
Eγ
⎛
⎜⎝ 2 + m c 2 ⎟⎠ cos θ = 1 + m c 2
e
e
which reduces to:
pγ′ =
[1]
[2]
2 cos θ − 1 =
or
or cos θ =
h
λ′
443
=
From energy conservation:
K e = Eγ − Eγ′ = 0.880 MeV − 0.602 MeV = 0.278 MeV = 278 keV
444
P40.26
Chapter 40
The energy of the incident photon is E0 = pγ c =
(a)
hc
.
λ0
Conserving momentum in the x direction gives
pλ = pe cos φ + pγ′ cos θ , or since φ = θ ,
(
)
E0
= pe + pγ′ cos θ
c
[1]
Conserving momentum in the y direction (with φ = θ ) yields
0 = pγ′ sin θ − pe sin θ , or pe = pγ′ =
h
λ′
[2]
Substituting Equation [2] into Equation [1] gives
2hc
E0 ⎛ h
h
cos θ
= ⎜ + ⎞⎟ cos θ , or λ ′ =
⎝
⎠
E0
λ′ λ′
c
h
By the Compton equation, λ ′ − λ0 =
(1 − cos θ ),
me c
2hc
2hc
h
cos θ −
=
(1 − cos θ )
E0
E0
me c
which reduces to
( 2m c
e
+ E0 ) cos θ = me c 2 + E0
From Equation [3],
λ′ =
2hc
2hc ⎛ me c 2 + E0 ⎞
cos θ =
E0
E0 ⎜⎝ 2 me c 2 + E0 ⎟⎠
Therefore,
Eγ′ =
hc
hc
=
2
λ ′ ( 2hc E0 ) ( me c + E0 ) ( 2 me c 2 + E0 )
=
and
(c)
2
⎛ m c2 + E ⎞
φ = θ = cos −1 ⎜ e 2 0 ⎟
⎝ 2 me c + E 0 ⎠
Thus,
(b)
[3]
pγ′ =
E 0 ⎛ 2 me c 2 + E 0 ⎞
2 ⎜⎝ me c 2 + E0 ⎟⎠
Eγ′
c
=
E 0 ⎛ 2 me c 2 + E 0 ⎞
2c ⎜⎝ me c 2 + E0 ⎟⎠
From conservation of energy, K e = E0 − Eγ′ = E0 −
or
Ke =
E0
2
Finally, from Equation (2), pe = pγ′ =
E0
2
⎛ 2 me c 2 + E 0 ⎞
⎜⎝ m c 2 + E ⎟⎠
e
0
⎛ 2 me c + 2 E 0 − 2 me c − E 0 ⎞
E02
=
⎟⎠
⎜⎝
me c 2 + E 0
2 ( me c 2 + E 0 )
2
E 0 ⎛ 2 me c 2 + E 0 ⎞
.
2c ⎜⎝ me c 2 + E0 ⎟⎠
2
Introduction to Quantum Physics
*P40.27 (a)
K=
1
me u 2 :
2
E ′ = E0 − K and
K=
1
(9.11 × 10 −31 kg) (1.40 × 106 m s )2 = 8.93 × 10 −19 J = 5.58 eV
2
E0 =
hc 1 240 eV ⋅ nm
=
= 1 550 eV
λ0
0.800 nm
λ′ =
hc
1 240 eV ⋅ nm
=
= 0.803 nm
E ′ 1 550 eV − 5.58 eV
∆λ = λ ′ − λ0 = 0.002 89 nm = 2.89 pm
(b)
P40.28
∆λ
0.002 89 nm
= 1−
= − 0.189
λC
0.002 43 nm
∆λ = λ C (1 − cos θ ) : cos θ = 1 −
θ = 101°
so
The electron’s kinetic energy is
K=
2
1
1
mu 2 = 9.11 × 10 −31 kg ( 2.18 × 10 6 m s ) = 2.16 × 10 −18 J
2
2
This is the energy lost by the photon, hf0 − hf ′
hc hc
−
= 2.16 × 10 −18 J. We also have
λ0 λ ′
λ ′ − λ0 =
h
6.63 × 10 −34 J ⋅ s ⋅ s
(1 − cos θ ) =
(1 − cos 17.4° )
me c
9.11 × 10 −31 kg ( 3 × 108 m )
λ ′ = λ0 + 1.11 × 10 −13 m
(a)
Combining the equations by substitution,
1
1
2.16 × 10 −18 J ⋅ s
−
=
= 1.09 × 10 7 m
λ0 λ0 + 0.111 pm 6.63 × 10 −34 J ⋅ s ( 3 × 108 m )
λ0 + 0.111 pm − λ0
= 1.09 × 10 7 m
λ02 + λ0 ( 0.111 pm )
0.111 pm = (1.09 × 10 7 m ) λ02 + 1.21 × 10 −6 λ0
1.09 × 10 7 λ02 + 1.21 × 10 −6 mλ0 − 1.11 × 10 −13 m 2 = 0
λ0 =
−1.21 × 10 −6 m ±
(1.21 × 10
−6
m ) − 4 (1.09 × 10 7 ) ( −1.11 × 10 −13 m 2 )
2
2 (1.09 × 10 7 )
only the positive answer is physical: λ0 = 1.01 × 10 −10 m .
(b)
445
Then λ ′ = 1.01 × 10 −10 m + 1.11 × 10 −13 m = 1.01 × 10 −10 m.
Conservation of momentum in the transverse direction:
h
0 = sin θ − γ me u sin φ
λ′
9.11 × 10 −31 kg ( 2.18 × 10 6 m s ) sin φ
6.63 × 10 −34 J ⋅ s
sin 17.4° =
2
1.01 × 10 −10 m
1 − ( 2.18 × 10 6 3 × 108 )
1.96 × 10 −24 = 1.99 × 10 −24 sin φ
φ = 81.1°
446
Chapter 40
*P40.29 It is, because Compton’s equation and the conservation of vector momentum give three
independent equations in the unknowns l⬘, l0, and u. They are
λ ′ − λ0 =
h
(1 − cos 90°)
me c
so λ ′ = λ0 +
h
me c
h
= γ me u cos 20°
λ0
h
= γ me u sin 20°
λ′
Dividing the latter two equations gives
λ0
= tan 20°
λ′
Then substituting, λ ′ = λ ′ tan 20° +
So l⬘ = 2.43 × 10–12 m/(1 – tan 20°) = 3.82 pm
P40.30
λ′ − λ =
h
(1 − cos θ )
me c
λ ′′ − λ ′ =
h
[1 − cos (π − θ )]
me c
λ ′′ − λ =
h
h
h
h
−
cos θ
−
cos (π − θ ) +
me c me c
me c me c
Now cos (π − θ ) = − cos θ , so
λ ′′ − λ = 2
P40.31
FIG. P40.30
h
= 0.004 86 nm
me c
Maximum energy loss appears as maximum increase in wavelength, which occurs for
h ⎞ 2h
=
scattering angle 180°. Then ∆λ = (1 − cos 180° ) ⎛⎜
where m is the mass of the
⎝ mc ⎟⎠ mc
target particle. The fractional energy loss is
E0 − E ′ hc λ0 − hc λ ′ λ ′ − λ0
∆λ
2h mc
=
=
=
=
E0
hc λ0
λ′
λ0 + ∆λ λ0 + 2h mc
Further, λ0 =
(a)
2 E0
E − E′
hc
2h mc
, so 0
.
=
=
E0
E0
hc E0 + 2h mc mc 2 + 2 E0
For scattering from a free electron, mc 2 = 0.511 MeV, so
E0 − E ′
2 ( 0.511 MeV )
= 0.667
=
0.511 MeV + 2 ( 0.511 MeV )
E0
(b)
For scattering from a free proton, mc 2 = 938 MeV, and
E0 − E ′
2 ( 0.511 MeV )
= 0.001 09
=
938 MeV + 2 ( 0.511 MeV )
E0
h
me c
Introduction to Quantum Physics
Section 40.4
447
Photons and Electromagnetic Waves
*P40.32 With photon energy 10.0 eV = hf a photon would have
f =
10.0 (1.6 × 10 −19 J )
6.63 × 10 −34 J ⋅ s
= 2.41 × 1015 Hz and l = c/f = (3 × 108 m/s)/(2.41 × 1015/s) = 124 nm
To have photon energy 10 eV or greater, according to this definition, ionizing radiation is the
ultraviolet light, x-rays, and g rays with wavelength shorter than 124 nm; that is, with frequency
higher than 2.41 × 1015 Hz.
P40.33
The photon energy is E =
−34
8
hc 6.63 × 10 J ⋅ s ( 3 × 10 m s )
=
= 3.14 × 10 −19 J. The power carried
λ
633 × 10 −9 m
by the beam is ( 2 × 1018 photons s ) ( 3.14 × 10 −19 J photon ) = 0.628 W. Its intensity is the
average Poynting vector I = Sav =
(a)
Sav =
P
πr
2
=
0.628 W ( 4 )
π (1.75 × 10 −3 m )
2
= 2.61 × 10 5 W m 2 .
E2
1
1 Emax Bmax
Erms Brms sin 90° =
. Also Emax = Bmax c . So Sav = max .
µ0
µ0 2 2
2µ0 c
Emax = ( 2 µ0 cSav )
12
(
= 2 ( 4π × 10 −7 Tm A ) ( 3 × 108 m s ) ( 2.61 × 10 5 W m 2 )
)
12
= 1.40 × 10 4 N C
Bmax =
(b)
1.40 × 10 4 N C
= 4.68 × 10 −5 T
3 × 108 m s
Each photon carries momentum
P
E
. The beam transports momentum at the rate .
c
c
It imparts momentum to a perfectly reflecting surface at the rate
2P
2 ( 0.628 W )
= force =
= 4.19 × 10 −9 N
c
3 × 108 m s
(c)
The block of ice absorbs energy mL = P ∆t melting
m=
P ∆t
L
=
0.628 W (1.5 × 3 600 s )
= 1.02 × 10 −2 kg
3.33 × 10 5 J kg
448
Chapter 40
Section 40.5
P40.34
λ=
P40.35
(a)
The Wave Properties of Particles
h
h
6.626 × 10 −34 J ⋅ s
=
=
= 3.97 × 10 −13 m
p mu (1.67 × 10 −27 kg ) (1.00 × 10 6 m s )
p2
= ( 50.0 ) (1.60 × 10 −19 J )
2m
p = 3.81 × 10 −24 kg ⋅ m s
h
λ = = 0.174 nm
p
(b)
p2
= ( 50.0 × 10 3 ) (1.60 × 10 −19 J )
2m
p = 1.20 × 10 −22 kg ⋅ m s
h
λ = = 5.49 × 10 −12 m
p
The relativistic answer is slightly more precise:
λ=
P40.36
(a)
h
hc
−12
=
m
1 2 = 5.37 × 10
p ⎡( mc 2 + K )2 − m 2 c 4 ⎤
⎣
⎦
Electron:
λ=
and
λ=
h
p
and
h
=
2 me K
K=
1
m 2u 2
p2
me u 2 = e =
2
2 me
2 me
6.626 × 10 −34 J ⋅ s
so
p = 2 me K
2 ( 9.11 × 10 −31 kg ) ( 3.00 ) (1.60 × 10 −19 J )
λ = 7.09 × 10 −10 m = 0.709 nm
(b)
*P40.37 (a)
Photon:
λ=
c
f
and
λ=
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
= 4.14 × 10 −7 m = 414 nm
E
3 (1.60 × 10 −19 J )
λ ~ 10 −14 m or less.
and
p=
The energy of the electron is E =
(b)
E = hf
so
f =
E
h
h 6.6 × 10 −34 J ⋅ s
~
= 10 −19 kg ⋅ m s or more
λ
10 −14 m
p 2 c 2 + me2 c 4 ~
(10 ) (3 × 10 ) + (9 × 10 ) (3 × 10 )
−19 2
8 2
−31 2
or
E ~ 10 −11 J ~ 108 eV or more
so that
K = E − me c 2 ~ 108 eV − ( 0.5 × 10 6 eV ) ~ 108 eV or more
If the nucleus contains ten protons, the electric potential energy of the electron-nucleus
system would be
Ue =
9
2
2
−19
ke q1q2 ( 9 × 10 N ⋅ m C ) (10 × 10 C ) ( − e )
~
5 × 10 −15 m
r
With its K + U e >> 0
~ −106 eV
the electron would immediately escape the nuucleus .
8 4
Introduction to Quantum Physics
P40.38
449
From the condition for Bragg reflection,
φ
mλ = 2d sin θ = 2d cos ⎛⎜ ⎞⎟
⎝ 2⎠
But
φ
d = a sin ⎛⎜ ⎞⎟
⎝ 2⎠
where a is the lattice spacing.
φ
φ
λ = 2a sin ⎛⎜ ⎞⎟ cos ⎛⎜ ⎞⎟ = a sin φ
⎝ 2⎠
⎝ 2⎠
Thus, with m = 1 ,
λ=
h
=
p
h
2 me K
6.626 × 10
λ=
2 ( 9.11 × 10
Therefore, the lattice spacing is a =
P40.39
(a )
(b)
−31
−34
FIG. P40.38
J⋅s
kg ) ( 54.0 × 1.60 × 10
−19
J)
= 1.67 × 10 −10 m
λ
1.67 × 10 −10 m
=
= 2.18 × 10 −10 = 0.218 nm .
sin φ
sin 50.0°
E 2 = p2 c2 + m 2 c4
h
λ
with E = hf ,
p=
so
h2 f 2 =
For a photon
f 1
=
c λ
The third term
h
λC
and
mc =
and
⎛ f⎞ = 1 + 1
⎜⎝ ⎟⎠
c
λ 2 λC2
2
h2c2 h2c2
+ 2
λ2
λC
(Eq. 1)
1
in Equation 1 for electrons and other massive particles shows
λC
that they will always have a different frequency from photons of the same wavelength .
*P40.40 (a)
For the massive particle, K = (γ − 1) mc 2 and λ m =
represent as g ), E = K and λγ =
λγ
chγ mu
γ u
=
=
2
λ m (γ − 1) mc h γ − 1 c
(
u
c 1 − 1 − u 2 /c 2
(b)
(c)
(d)
(e)
(f)
h
h
. For the photon (which we
=
p γ mu
c ch ch
ch
. Then the ratio is
=
=
=
f
E
K (γ − 1) mc 2
where γ =
1
1 − u 2 /c 2
. The ratio can be written
)
λγ
1 ( 0.9 )
=
= 1.60
λm
1 − 0.92 ⎡ 1 1 − 0.92 − 1⎤
⎣
⎦
The ratio for a particular particle speed does not depend on the particle mass: There would
be no change.
λγ
1 ( 0.001)
= 2.00 × 10 3
=
2
2
λm
⎡
⎤
1 − ( 0.001) ⎢ 1 1 − ( 0.001) − 1⎥
⎣
⎦
λγ
u
γ
→ 1= 1 .
As → 1, γ → ∞ and γ − 1 becomes nearly equal to g. Then
c
λm
γ
2 −1 2
2
2
u
λ
⎛
1 u
1u
u ⎞
2c
uc
As → 0 , ⎜ 1 − 2 ⎟ − 1 ≈ 1 − ⎛⎜ − ⎞⎟ 2 − 1 =
and γ → 1
=
→ ∞ .
2
2
c
⎝ 2⎠ c
⎝
2 c2
c ⎠
λm
(1 2) ( u c ) u
(
)
(
)
450
P40.41
Chapter 40
λ=
h
p
p=
h 6.626 × 10 −34 J ⋅ s
=
= 6.63 × 10 −23 kg ⋅ m s
λ
1.00 × 10 −11 m
(6.63 × 10 −23 ) J = 15.1 keV
p2
Ke =
=
2 me 2 ( 9.11 × 10 −31 )
2
(a)
electrons:
The relativistic answer is more precisely correct:
Ke =
(b)
*P40.42 (a)
Eγ = pc = ( 6.63 × 10 −23 ) ( 3.00 × 108 ) = 124 keV
photons:
h
h
=
. If w is the width of the diffracting aperture,
p mu
h ⎞
w ≤ 10.0 λ = 10.0 ⎛⎜
⎝ mu ⎟⎠
The wavelength of the student is λ =
then we need
u ≤ 10.0
so that
(b)
p 2 c 2 + me2 c 4 − me c 2 = 14.9 keV
Using ∆t =
d
we get:
u
⎛ 6.626 × 10 −34 J ⋅ s ⎞
h
= 1.10 × 10 −34 m s
= 10.0 ⎜
mw
⎝ (80.0 kg ) ( 0.750 m ) ⎟⎠
∆t ≥
0.150 m
= 1.36 × 10 33 s , which is more
1.10 × 10 −34 m s
than 1015 times the age of the Universe.
(c)
*P40.43 (a)
(b)
He should not worry. The student cannot be diffracted to a measurable extent. Even if
he tries to stand still, molecular bombardment will give him a sufficiently high speed to
make his wavelength immeasurably small.
E = g mc2
g = E/mc2 = 20 000 MeV/0.511 MeV = 3.91 × 104
pc = [E 2 – (mc2)2]1/ 2 = [(20 000 MeV)2 – (0.511 MeV)2]1/ 2 = 20.0 GeV
p = 20.0 GeV/c = 1.07 × 10–17 kg ⋅ m/s
(c)
Section 40.6
P40.44
l = h/p = 6.626 × 10–34 J ⋅ s /1.07 × 10–17 kg ⋅ m/s = 6.21 × 10–17 m , small compared to the
size of the nucleus. The scattering of the electrons can give information about the particles
forming the nucleus.
The Quantum Particle
E=K=
1
h
mu 2 = hf and λ =
2
mu
vphase = f λ =
mu 2 h
u
=
= vphase
2h mu
2
This is different from the speed u at which the particle transports mass, energy, and momentum.
Introduction to Quantum Physics
P40.45
As a bonus, we begin by proving that the phase speed v p =
vp =
ω
=
k
= c 1+
ω
is not the speed of the particle.
k
γ 2 m 2u 2 c2 + m 2 c4
p2 c2 + m 2 c4 c2
c2 ⎛
u2 ⎞
=
=
c
1
+
=
1
+
1
−
c
2
2
2
⎜
γ u
γ mu
u ⎝
c 2 ⎟⎠
γ 2 m 2u 2
c2
c2
−1 =
2
u
u
In fact, the phase speed is larger than the speed of light. A point of constant phase in the wave
function carries no mass, no energy, and no information.
Now for the group speed:
vg =
dω d ω dE
d
=
=
=
m 2 c4 + p2 c2
dk
d k dp dp
vg =
1 2 4
( m c + p 2 c2 )−1 2 ( 0 + 2 pc2 ) =
2
vg = c
p2 c4
p c + m 2 c4
2 2
u 2 (1 − u 2 c 2 )
γ 2 m 2u 2
=c
=
c
γ 2 m 2u 2 + m 2 c2
u 2 (1 − u 2 c 2 ) + c 2
(u
u 2 (1 − u 2 c 2 )
2
+ c 2 − u 2 ) (1 − u 2 c 2 )
=u
It is this speed at which mass, energy, and momentum are transported.
Section 40.7
P40.46
The Double-Slit Experiment Revisited
Consider the first bright band away from the center:
d sin θ = mλ
λ=
h
me u
so
and
h2
∆V =
2eme λ 2
(6.00 × 10
me u =
K=
−8
0.400 ⎤⎞
⎛
−10
m ) sin ⎜ tan −1 ⎡⎢
⎟ = (1) λ = 1.20 × 10 m
⎝
⎣ 200 ⎦⎥⎠
h
λ
1
m 2u 2
h2
me u 2 = e =
= e∆V
2
2 me
2 me λ 2
(6.626 × 10
C ) ( 9.11 × 10
−34
∆V =
2 (1.60 × 10 −19
−31
J ⋅ s)
2
kg ) (1.20 × 10 −10 m )
2
= 105 V
451
452
P40.47
Chapter 40
(a)
(b)
λ=
h
6.626 × 10 −34 J ⋅ s
=
= 9.92 × 10 −7 m
mu (1.67 × 10 −27 kg ) ( 0.400 m s )
1
For destructive interference in a multiple-slit experiment, d sin θ = ⎛⎜ m + ⎞⎟ λ , with m = 0
⎝
2⎠
for the first minimum. Then
⎛ λ⎞
θ = sin −1 ⎜ ⎟ = 0.028 4°
⎝ 2d ⎠
y
= tan θ
L
(c)
P40.48
y = L tan θ = (10.0 m ) ( tan 0.028 4° ) = 4.96 mm
We cannot say the neutron passed through one slit. If its detection forms part of an
interference pattern, we can only say it passed through the array of slits. If we test to see
which slit a particular neutron passes through, it will not form part of the interference
pattern.
We find the speed of each electron from energy conservation in the firing process:
0 = Kf +Uf =
u=
2eV
=
m
1
mu 2 − eV
2
2 × 1.6 × 10 −19 C ( 45 V )
= 3.98 × 10 6 m s
9.11 × 10 −31 kg
The time of flight is ∆t =
28 cm apart is I =
Section 40.8
P40.49
q
e 1.6 × 10 −19 C
=
=
= 2.27 × 10 −12 A .
t ∆t 7.04 × 10 −8 s
The Uncertainty Principle
For the electron,
∆p = me ∆u = ( 9.11 × 10 −31 kg ) ( 500 m s ) (1.00 × 10 −4 ) = 4.56 × 10 −32 kg ⋅ m s
∆x =
For the bullet,
6.626 × 10 −34 J ⋅ s
h
=
= 1.16 mm
4π ∆p 4π ( 4.56 × 10 −32 kg ⋅ m s )
∆p = m ∆u = ( 0.020 0 kg )( 500 m s ) (1.00 × 10 −4 ) = 1.00 × 10 −3 kg ⋅ m s
∆x =
P40.50
0.28 m
∆x
=
= 7.04 × 10 −8 s. The current when electrons are
3.98 × 10 6 m s
u
h
= 5.28 × 10 −32 m
4π ∆p
2
∆p∆ x = m ∆u ∆ x ≥
(b)
The duck might move by ( 0.25 m s ) ( 5 s ) = 1.25 m. With original position uncertainty of
so
∆u ≥
h
2π J ⋅ s
=
= 0.250 m s
4π m ∆x 4π ( 2.00 kg ) (1.00 m )
(a)
1.00 m, we can think of ∆ x growing to 1.00 m + 1.25 m = 2.25 m .
Introduction to Quantum Physics
P40.51
∆y ∆py
=
x
px
d ∆py ≥
and
453
h
4π
Eliminate ∆py and solve for x.
d
x = 4π px ( ∆y ) :
h
x = 4π (1.00 × 10 −3 kg ) (100 m s ) (1.00 × 10 −2 m )
( 2.00 × 10
(6.626 × 10
−3
−34
m)
J ⋅ s)
The answer, x = 3.79 × 10 28 m , is 190 times greater than the diameter of the observable Universe.
P40.52
= 2.6 × 10 −20 kg ⋅ m s .
2 ∆x
The average momentum of the particle bound in a stationary nucleus is zero. The uncertainty in
momentum measures the root-mean-square momentum, so we take prms ≈ 3 × 10 −20 kg ⋅ m s. For
With ∆x = 2 × 10 −15 m, the uncertainty principle requires ∆px ≥
an electron, the non-relativistic approximation p = me u would predict u ≈ 3 × 1010 m s, while u
cannot be greater than c.
2
2
E = ⎡( me c 2 ) + ( pc ) ⎤
⎣
⎦
12
Thus, a better solution would be
1
γ ≈ 110 =
For a proton, u =
P40.53
(a)
≈ 56 MeV = γ me c 2
so
1 − v2 c2
u ≈ 0.999 96c
p
gives u = 1.8 × 10 7 m s, less than one-tenth the speed of light.
m
At the top of the ladder, the woman holds a pellet inside a small region ∆ xi. Thus, the
uncertainty principle requires her to release it with typical horizontal momentum
1
∆px = m ∆u x =
. It falls to the floor in a travel time given by H = 0 + gt 2 as
2 ∆xi
2
2H
, so the total width of the impact points is
g
t=
⎛ ⎞
∆ x f = ∆xi + ( ∆u x ) t = ∆xi + ⎜
⎝ 2 m ∆xi ⎟⎠
A=
where
To minimize ∆x f , we require
2H
2m g
( )=0
d ∆x f
d ( ∆xi )
or
1−
A
=0
∆xi2
∆xi = A
so
The minimum width of the impact points is
( ∆x )
f
(b)
( )
∆x f
min
⎛
A ⎞
= ⎜ ∆xi +
∆xi ⎟⎠
⎝
=2 A=
∆xi = A
⎡ 2 (1.054 6 × 10 −34 J ⋅ s ) ⎤
=⎢
⎥
5.00 × 10 −4 kg
⎥⎦
⎣⎢
12
min
2 ⎛ 2 H ⎞
m ⎜⎝ g ⎠⎟
⎡ 2 ( 2.00 m ) ⎤
⎢ 9.80 m s 2 ⎥
⎣
⎦
14
14
= 5.19 × 10 −16 m
2H
A
= ∆xi +
g
∆xi
454
Chapter 40
Additional Problems
*P40.54 The condition on electric power delivered to the filament is
P = I ∆V =
( ∆V )2
R
=
( ∆V )2 A
ρ
=
( ∆V )2 π r 2
ρ
⎛ P ρ ⎞
so r = ⎜
2
⎝ π ( ∆V ) ⎟⎠
12
. Here
P = 75 W,
ρ = 7.13 × 10 −7 Ω ⋅ m , and ∆V = 120 V. As the filament radiates in steady state, it must emit
all of this power through its lateral surface area
conditions by substitution:
⎛ P ρ ⎞
2
⎝ π ( ∆V ) ⎟⎠
P = σ eAT 4 = σ e2π r T 4. We combine the
12
P = σ e2π ⎜
T 4
( ∆V ) P 1 2 = σ e2π 1 2 ρ1 2 3 2T 4
⎛ ( ∆V ) P 1 2 ⎞
=⎜
⎝ σ e2π 1 2 ρ1 2T 4 ⎟⎠
23
12
⎛
⎞
120 V ( 75 W ) m 2 K 4
=⎜
1
2
4⎟
−8
12
−7
⎝ 5.67 × 10 W 0.450 ( 2 ) π ( 7.13 × 10 Ω ⋅ m ) ( 2 900 K ) ⎠
= ( 0.192 m 3 2 )
⎛ P ρ ⎞
r=⎜
2
⎝ π ( ∆V ) ⎟⎠
and
P40.55
12
23
23
= 0.3333 m = ⎛ 75 W 7.13 × 10 −7 Ω ⋅ m 0.333 m ⎞
=⎜
⎟⎠
π (120 V )2
⎝
12
= r = 1.98 × 10 −5 m
We want an Einstein plot of K max versus f
λ , nm
588
505
445
399
f , 1014 Hz K max , eV
5.10
0.67
5.94
0.98
6.74
1.35
7.52
1.63
0.402 eV
± 8%
1014 Hz
(a)
slope =
(b)
e∆VS = hf − φ
⎛ 1.60 × 10 −19 J ⋅ s ⎞
−34
h = ( 0.402 ) ⎜
⎟⎠ = 6.4 × 10 J ⋅ s ± 8%
⎝
1014
(c)
K max = 0
at f ≈ 344 × 1012 Hz
φ = hf = 2.32 × 10 −19 J = 1.4 eV
FIG. P40.55
Introduction to Quantum Physics
P40.56
h
φ
∆VS = ⎛⎜ ⎞⎟ f −
⎝ e⎠
e
From two points on the graph
h
φ
0 = ⎛⎜ ⎞⎟ ( 4.1 × 1014 Hz ) −
⎝ e⎠
e
and
φ
h
3.3 V = ⎛⎜ ⎞⎟ (12 × 1014 Hz ) −
⎝ e⎠
e
Combining these two expressions we find:
(a)
φ = 1.7 eV
(b)
h
= 4.2 × 10 −15 V ⋅ s
e
(c)
At the cutoff wavelength
FIG. P40.56
hc
h ec
= φ = ⎛⎜ ⎞⎟
⎝
λc
e ⎠ λc
λc = ( 4.2 × 10 −15 V ⋅ s ) (1.6 × 10 −19 C )
P40.57
455
(3 × 10 m s)
=
(1.7 eV ) (1.6 × 10 J eV )
8
−19
730 nm
From the path the electrons follow in the magnetic field, the maximum kinetic energy is seen to be:
K max =
e2 B 2 R 2
2 me
From the photoelectric equation,
K max = hf − φ =
Thus, the work function is
φ=
*P40.58 (a)
hc
−φ
λ
hc
hc e2 B 2 R 2
− K max =
−
2 me
λ
λ
We find the energy of one photon:
2
⎛ 1.6 × 10 −19 J ⎞ 1
−31
3
−19
−19
3.44 eV ⎜
⎟⎠ + 2 9.11 × 10 kg ( 420 × 10 m/s ) = 5.50 × 10 J + 0.804 × 10 J
⎝
1 eV
= 6.31 × 10 −19 J
The number intensity of photon bombardment is
550 J/s ⋅ m 2 ⎛ 1 m 2 ⎞
= 8.72 × 1016 1/s ⋅ cm 2
6.31 × 10 −19 J ⎜⎝ 10 4 cm 2 ⎟⎠
(b)
The density of the current the imagined electrons comprise is
8.72 × 1016
(c)
1
1.6 × 10 −19 C = 14.0 mA/cm 2
s ⋅ cm 2
Many photons are likely reflected or give their energy to the metal as internal energy, so the
actual current is probably a small fraction of 14.0 mA.
456
P40.59
Chapter 40
Isolate the terms involving ø in Equations 40.13 and 40.14. Square and add to eliminate ø.
⎡1
1
2 cos θ ⎤
2 2 2
h2 ⎢ 2 + 2 −
⎥ = γ me u
λ
λ
λ
λ
′
′
0
⎣ 0
⎦
Solve for
1
2 cos θ ⎤
h2 ⎡ 1
u2
b
b
=
+ 2−
where
we
defi
ne
=
2 ⎢ 2
2
2
λ
λ
λ0 λ ′ ⎥⎦
m
′
c
(b + c )
e ⎣ 0
Substitute into Eq. 40.12:
−1 2
⎛ h ⎞⎡ 1
1⎤
b ⎞
⎛
1+ ⎜
=
⎢ − ⎥ = γ = ⎜⎝ 1 −
2⎟
⎟
b+c ⎠
⎝ me c ⎠ ⎣ λ 0 λ ′ ⎦
Square each side:
2hc ⎡ 1
1 ⎤ h2 ⎡ 1
1⎤
c +
⎢ − ⎥+ 2 ⎢ − ⎥
me ⎣ λ 0 λ ′ ⎦ me ⎣ λ 0 λ ′ ⎦
c2 + b
c2
2
2
⎛ h2 ⎞ ⎡ 1
1
2 cos θ ⎤
= c2 + ⎜ 2 ⎟ ⎢ 2 + 2 −
λ0 λ ′ ⎥⎦
⎝ me ⎠ ⎣ λ 0 λ ′
From this we get Eq. 40.11:
P40.60
⎛ h ⎞
λ ′ − λ0 = ⎜
[1 − cos θ ]
⎝ me c ⎟⎠
We show that if all of the energy of a photon is transmitted to an electron, momentum will not be
conserved.
Energy:
hc hc
=
+ K e = me c 2 (γ − 1)
λ0 λ ′
Momentum:
h
h
=
+ γ me u = γ me u
λ0 λ ′
From (1),
γ =
if
if
hc
=0
λ′
(1)
λ′ = ∞
(2)
h
+1
λ 0 me c
(3)
⎛ λ 0 me c ⎞
u = c 1− ⎜
⎝ h + λ0 me c ⎟⎠
2
(4)
Substitute (3) and (4) into (2) and show the inconsistency:
⎛ λ 0 me c ⎞
λ m c + h h ( h + 2 λ 0 me c ) h
h ⎛
h ⎞
= 0 e
me c 1 − ⎜
= ⎜1 +
=
⎟
⎟
λ0 ⎝
λ 0 me c ⎠
λ0
⎝ h + λ 0 me c ⎠
( h + λ 0 m e c )2 λ 0
2
h + 2 λ 0 me c
h
This is impossible, so all of the energy of a photon cannot be transmitted to an electron.
Introduction to Quantum Physics
P40.61
(a)
I (λ, T ) =
Starting with Planck’s law,
2π hc 2
λ 5 ⎡⎣ ehc λ kBT − 1⎤⎦
∞
the total power radiated per unit area
∫ I (λ, T ) dλ =
0
∞
2π hc 2
∫0 λ 5 ⎡ ehc λkBT − 1⎤ d λ
⎣
⎦
hc
λ kBT
Change variables by letting
x=
and
dx = −
hcd λ
kBT λ 2
Note that as l varies from 0 → ∞ , x varies from
∞
∞ → 0.
∫ I (λ, T ) dλ = −
Then
457
0
∞
2π kB4T 4
h3c2
2π kB4T 4 ⎛ π 4 ⎞
x3
=
dx
∫ ex − 1)
h 3 c 2 ⎜⎝ 15 ⎟⎠
∞ (
0
⎛ 2π 5 kB4 ⎞ 4
T =σT4
3 2⎟
⎠
∫ I ( λ , T ) d λ = ⎜⎝ 15h c
Therefore,
0
2π 5 (1.38 × 10 −23 J K )
2π 5 kB4
=
3
2
15h 3 c 2 15 ( 6.626 × 10 −34 J ⋅ s ) ( 3.00 × 108 m s )
4
(b)
σ=
From part (a),
σ = 5.67 × 10 −8 W m 2 ⋅ K 4
P40.62
Planck’s law states I ( λ , T ) =
−1
2π hc 2
= 2π hc 2 λ −5 ⎡⎣ ehc λ kBT − 1⎤⎦ .
λ ⎡⎣ ehc λ kBT − 1⎤⎦
5
To find the wavelength at which this distribution has a maximum, compute
⎧⎪
−2
−1
dI
= 2π hc 2 ⎨−5λ −6 ⎡⎣ ehc λ kBT − 1⎤⎦ − λ −5 ⎡⎣ ehc λ kBT − 1⎤⎦ ehc λ kBT
dλ
⎪⎩
⎛
hc ⎞ ⎫⎪
⎜⎝ − λ 2 k T ⎟⎠ ⎬ = 0
⎪⎭
B
⎧⎪
2π hc 2
dI
hc
ehc λ kBT ⎫⎪
−5 +
= 6 hc λ k T
⎬=0
⎨
d λ λ ⎡⎣ e B − 1⎤⎦ ⎩⎪
λ kBT ⎡⎣ ehc λ kBT − 1⎤⎦ ⎭⎪
Letting x =
xe x
hc
= 5.
, the condition for a maximum becomes x
λ kBT
e −1
We zero in on the solution to this transcendental equation by iterations as shown in the table
below.
continued on next page
458
Chapter 40
x
xe x ( e x − 1)
x
xe x ( e x − 1)
4.000 00
4.074 629 4
4.964 50
4.999 403 0
4.500 00
4.550 552 1
4.965 50
5.000 374 9
5.000 00
5.033 918 3
4.965 00
4.999 889 0
4.900 00
4.936 762 0
4.965 25
5.000 132 0
4.950 00
4.985 313 0
4.965 13
5.000 015 3
4.975 00
5.009 609 0
4.965 07
4.999 957 0
4.963 00
4.997 945 2
4.965 10
4.999 986 2
4.969 00
5.003 776 7
4.965 115
5.000 000 8
4.966 00
5.000 860 9
The solution is found to be
hc
x=
= 4.965 115 and
λmax kBT
Thus, λmaxT =
(6.626 075 × 10
λmaxT =
−34
hc
4.965 115kB
J ⋅ s ) ( 2.997 925 × 108 m s )
4.965 115 (1.380 658 × 10 −23 J K )
= 2.897 755 × 10 −3 m ⋅ K .
This result agrees with Wien’s experimental value of λmaxT = 2.898 × 10 −3 m ⋅ K for this
constant.
P40.63
p = mu = 2 mE = 2 (1.67 × 10 −27 kg ) ( 0.040 0 eV ) (1.60 × 10 −19 J eV )
λ=
h
= 1.43 × 10 −10 m = 0.143 nm
mu
This is of the same order of magnitude as the spacing between atoms in a crystal, so diffraction
should appear. A diffraction pattern with maxima and minima at the same angles can be
produced with x-rays, with neutrons, and with electrons of much higher kinetic energy, by using
incident quantum particles with the same wavelength.
P40.64
(a)
mgyi =
1
mu 2f
2
u f = 2 gyi = 2 ( 9.80 m s 2 ) ( 50.0 m ) = 31.3 m s
λ=
(b)
(c)
h
6.626 × 10 −34 J ⋅ s
= 2.82 × 10 −37 m
=
mu ( 75.0 kg )( 31.3 m s )
∆E ∆t ≥
2
so ∆E ≥
6.626 × 10 −34 J ⋅ s
= 1.06 × 10 −32 J
4π ( 5.00 × 10 −3 s )
( not observable )
∆E
1.06 × 10 −32 J
=
= 2.87 × 10 −35%
E
( 75.0 kg) ( 9.80 m s2 ) ( 50.0 m )
Introduction to Quantum Physics
P40.65
λC =
λ C h me c
p
=
=
h p
me c
λ
h
h
and λ = :
me c
p
E 2 = c 2 p 2 + ( me c 2 ) :
2
λC
1
E2
2
=
− ( me c ) =
2
λ me c c
P40.66
p=
E2
2
− ( me c )
2
c
2
1
( m e c )2
⎛ E ⎞
⎡ E2
2⎤
⎢ c 2 − ( me c ) ⎥ = ⎜⎝ m c 2 ⎟⎠ − 1
⎣
⎦
e
∆λ =
h
(1 − cos θ ) = λ ′ − λ0
me c
E′ =
⎡
⎤
hc
hc
h
=
= hc ⎢ λ0 +
(1 − cos θ ) ⎥
λ ′ λ0 + ∆λ
me c
⎣
⎦
⎤
hc ⎡
hc
E′ =
(1 − cos θ ) ⎥
⎢1 +
2
λ 0 ⎣ me c λ 0
⎦
−1
−1
−1
E′ =
P40.67
⎤
⎤
⎡
E
hc ⎡
hc
1+
(1 − cos θ ) ⎥ = E0 ⎢1 + 0 2 (1 − cos θ ) ⎥
m
c
λ0 ⎢⎣ me c 2 λ0
⎣
⎦
e
⎦
−1
2
From the uncertainty principle
∆E ∆t ≥
or
∆ ( mc 2 ) ∆t =
Therefore,
∆m
h
h
=
=
2
m
4π c ( ∆t ) m 4π ( ∆t ) ER
2
∆m
6.626 × 10 −34 J ⋅ s
⎛ 1 MeV ⎞
=
−17
m
4π (8.70 × 10 s ) (135 MeV ) ⎝ 1.60 × 10 −13 J ⎠
= 2.81 × 10 −8
459
460
P40.68
Chapter 40
Let u ′ represent the final speed of the electron and
⎛
u′2 ⎞
let γ ′ = ⎜ 1 − 2 ⎟
⎝
c ⎠
−1 2
. We must eliminate b and u ′
from the three conservation equations:
hc
hc
+ γ me c 2 =
+ γ ′ me c 2
λ0
λ′
[1]
h
h
+ γ me u − cos θ = γ ′ me u ′ cos β
λ0
λ′
[2]
h
sin θ = γ ′ me u ′ sin β
λ′
[3]
FIG. P40.68
Square Equations [2] and [3] and add:
h2
h 2 2hγ me u 2h 2 cos θ 2hγ me u cos θ
2 2 2
m
u
+
+
+
−
−
γ
= γ ′ 2 me2 u ′ 2
e
λ02
λ ′2
λ0
λ0 λ ′
λ′
h2 h2
2hγ me u 2hγ me u cos θ 2h 2 cos θ
me2 u ′ 2
−
−
=
+ 2 + γ 2 me2 u 2 +
2
λ0
λ′
λ0 λ ′
1 − u ′2 c2
λ0 λ ′
bu ′ 2
b
c2b
= me2 u ′ 2 and u ′ 2 = 2
.
= 2 2
2
2
c
me + b c
me c + b
Now square Equation [1] and substitute to eliminate γ ′ :
Call the left-hand side b. Then b −
h2
h 2 2hγ me c 2h 2 2hγ me c
me2 c 2
2 2 2
= me2 c 2 + b
m
c
+
+
+
−
−
=
γ
e
λ02
λ ′2
λ0
λ0 λ ′
λ′
1 − u ′2 c2
2
2
2
So we have h + h + γ 2 me2 c 2 + 2hγ me c − 2hγ me c − 2h
2
2
λ0 λ ′
λ0
λ′
λ0 λ ′
= me c 2 +
2hγ me u 2hγ me u coss θ 2h 2 cos θ
h2 h2
+ 2 + γ 2 me2 u 2 +
−
−
2
λ0 λ ′
λ0
λ′
λ0 λ ′
Multiply through by λ0 λ ′
me2 c 2
λ0 λ ′γ 2 +
λ λ ′γ 2 u 2 2hλ ′uγ 2hγλ0 u cos θ 2h 2 cos θ
2hλ ′γ 2hλ0γ
2h 2
+
−
−
−
− 2 2 = λ0 λ ′ + 0
me c
me c
me c
me c 2
me c 2
me c 2
me2 c 2
⎛
γ 2 u 2 ⎞ 2hγλ ′ ⎛ u ⎞ 2hγλ0 ⎛ u cos θ ⎞ 2h 2
λ0 λ ′ ⎜ γ 2 − 1 − 2 ⎟ +
(1 − cos θ )
⎜ 1 − ⎟⎠ =
⎜1 −
⎟+
⎝
c ⎠
me c ⎝
c
me c ⎝
c ⎠ me2 c 2
The first term is zero. Then
⎛ 1 − ( u cos θ ) c ⎞ hγ −1 ⎛ 1 ⎞
λ ′ = λ0 ⎜
⎟⎠ + m c ⎜⎝ 1 − u c ⎠⎟ (1 − cos θ )
⎝
1− u c
e
Since
u
u
u
γ −1 = 1 − ⎛⎜ ⎞⎟ = ⎛⎜ 1 − ⎞⎟ ⎛⎜ 1 + ⎞⎟
⎝
⎝ c⎠
c⎠ ⎝
c⎠
2
this result may be written as
⎛ 1 − ( u cos θ ) c ⎞
h 1+ u c
λ ′ = λ0 ⎜
+
(1 − cos θ )
⎟
⎝
⎠ me c 1 − u c
1− u c
It shows a specific combination of what looks like a Doppler shift and a Compton shift. This
problem is about the same as the first problem in Albert Messiah’s graduate text on quantum
mechanics.
Introduction to Quantum Physics
461
ANSWERS TO EVEN PROBLEMS
P40.2
(a) 999 nm (b) The wavelength emitted most strongly is infrared, and much more energy is
radiated at wavelengths longer than lmax than at shorter wavelengths.
P40.4
(a) ~10 −7 m ultraviolet
P40.6
(a) 70.9 kW
(b) ~10 −10 m gamma ray
(c) 7.99 × 1010 W m
(b) 580 nm
(e) 1.00 × 10 −227 W m
(i) 2.60 × 10 −9 W m
(f) 5.44 × 1010 W m
(d) 9.42 × 10 −1226 W m
(g) 7.38 × 1010 W m
(h) 0.260 W m
(j) 20 kW
P40.8
5.71 × 10 3 photons
P40.10
1.34 × 10 31
P40.12
(a) 0.263 kg (b) 1.81 W (c) –0.015 3°C s = –0.919 °C min (d) 9.89 mm (e) 2.01 × 10–20 J
(f) 8.98 × 1019 photon s
P40.14
(a) 296 nm, 1.01 PHz
P40.16
(a) 1.38 eV
P40.18
148 d, the classical theory is a gross failure
P40.20
(a) The incident photons are Doppler shifted to higher frequencies, and hence, higher energy.
(b) 3.87 eV
P40.22
(b) 2.71 V
(b) 334 THz
(c) 8.78 eV
(a) and (b) q, degrees
(c)
0
30
60
90
120
150
180
l⬘, pm
120.0
120.3
121.2
122.4
123.6
124.5
124.8
Ke , eV
0
27.9
104
205
305
376
402
180°. We could answer like this: The photon imparts the greatest momentum to the
originally stationary electron in a head-on collision. Here the photon recoils straight back
and the electron has maximum kinetic energy.
P40.24
(a) 488 fm
P40.26
⎛ m c2 + E ⎞
(a) cos −1 ⎜ e 2 0 ⎟
⎝ 2 me c + E 0 ⎠
(c) K e =
(b) 268 keV
(c) 31.5 keV
(b) Eγ′ =
E 0 ⎛ 2 me c 2 + E 0 ⎞
E 0 ⎛ 2 me c 2 + E 0 ⎞
p
=
′
,
γ
2 ⎜⎝ me c 2 + E0 ⎟⎠
2c ⎜⎝ me c 2 + E0 ⎟⎠
E02
E 0 ⎛ 2 me c 2 + E 0 ⎞
,
p
=
e
2c ⎜⎝ me c 2 + E0 ⎟⎠
2 ( me c 2 + E 0 )
P40.28
(a) 0.101 nm (b) 81.1°
P40.30
0.004 86 nm
P40.32
To have photon energy 10 eV or greater, according to this definition, ionizing radiation is the
ultraviolet light, x-rays, and g rays with wavelength shorter than 124 nm; that is, with frequency
higher than 2.41 × 1015 Hz.
462
Chapter 40
P40.34
397 fm
P40.36
(a) 0.709 nm
P40.38
0.218 nm
P40.40
(a)
P40.42
(a) 1.10 × 10–34 m|s (b) 1.36 × 1033 s is more than 1015 times the age of the Universe. (c) The
student cannot be diffracted to a measurable extent. Even if he tries to stand still, molecular
bombardment will give him a sufficiently high speed to make his wavelength immeasurably small.
P40.44
vphase = u/2
P40.46
105 V
P40.48
2.27 pA
P40.50
(a) 0.250 m s
P40.52
The electron energy must be ~100 mc2 or larger. The proton energy can be as small as 1.001 mc2,
which is within the range well described classically.
P40.54
length 0.333 m, radius 19.8 mm
P40.56
(a) 1.7 eV
P40.58
(a) 8.72 × 1016/s (b) 14.0 mA (c) Many photons are likely reflected or give their energy to the
metal as internal energy, so the actual current is probably a small fraction of 14.0 mA.
P40.60
See the solution.
P40.62
See the solution.
P40.64
(a) 2.82 × 10 −37 m
P40.66
See the solution.
P40.68
See the solution.
(b) 414 nm
γ u
where g =
γ −1 c
1
1 − u 2 /c 2
(b) 1.60
(c) no change
(d) 2.00 × 103
(e) 1
(f) ∞
(b) 2.25 m
−15
(b) 4.2 × 10 V ⋅ s
(c) 730 nm
(b) 1.06 × 10 −32 J
(c) 2.87 × 10 −35%
41
Quantum Mechanics
In chapters 39, 40, and 41 we use u to represent the speed of a particle with mass, reserving v for the
speeds associated with reference frames, wave functions, and photons.
Note:
ANSWERS TO QUESTIONS
CHAPTER OUTLINE
41.1
41.2
41.3
41.4
41.5
41.6
41.7
An Interpretation of Quantum
Mechanics
The Quantum Particle under
Boundary Conditions
The Schrödinger Equation
A Particle in a Well of Finite
Height
Tunneling Through a Potential
Energy Barrier
Applications of Tunneling
The Simple Harmonic Oscillator
Q41.1
A particle’s wave function represents its state, containing all the information there is about its location and
motion. The squared absolute value of its wave function
tells where we would classically think of the particle as
2
spending most its time. Ψ is the probability distribution
function for the position of the particle.
*Q41.2 For the squared wave function to be the probability per
length of finding the particle, we require
0.48
0.16
2
=
and ψ = 0.4/ nm
ψ =
7 nm − 4 nm nm
(i) Answer (e). (ii) Answer (e).
*Q41.3 (i) For a photon a and b are true, c false, d, e, f, and g true, h false, i and j true.
(ii) For an electron a is true, b false, c, d, e, f true, g false, h, i and j true.
Note that statements a, d, e, f, i, and j are true for both.
*Q41.4 We consider the quantity h2n2/8mL2.
In (a) it is h21/8m1(3 nm)2 = h2/72 m1 nm2.
In (b) it is h24/8m1(3 nm)2 = h2/18 m1 nm2.
In (c) it is h21/16m1(3 nm)2 = h2/144 m1 nm2.
In (d) it is h21/8m1(6 nm)2 = h2/288 m1 nm2.
In (e) it is 021/8m1(3 nm)2 = 0.
The ranking is then b > a > c > d > e.
Q41.5
The motion of the quantum particle does not consist of moving through successive
points. The particle has no definite position. It can sometimes be found on one side of a node and
sometimes on the other side, but never at the node itself. There is no contradiction here, for the
quantum particle is moving as a wave. It is not a classical particle. In particular, the particle does
not speed up to infinite speed to cross the node.
463
464
Chapter 41
Q41.6
Consider a particle bound to a restricted region of space. If its minimum energy were zero,
then the particle could have zero momentum and zero uncertainty in its momentum. At the same
time, the uncertainty in its position would not be infinite, but equal to the width of the region. In
such a case, the uncertainty product ∆ x ∆ px would be zero, violating the uncertainty principle.
This contradiction proves that the minimum energy of the particle is not zero.
*Q41.7 Compare Figures 41.4 and 41.7 in the text. In the square well with infinitely high walls,
the particle’s simplest wave function has strict nodes separated by the length L of the well. The
h
p2
h2
particle’s wavelength is 2L, its momentum
, and its energy
. Now in the well with
=
2L
2 m 8 mL2
walls of only finite height, the wave function has nonzero amplitude at the walls. In this finite-depth
well …
(i) The particle’s wavelength is longer, answer (a).
(ii) The particle’s momentum in its ground state is smaller, answer (b).
(iii) The particle has less energy, answer (b).
Q41.8
As Newton’s laws are the rules which a particle of large mass follows in its motion, so the
Schrödinger equation describes the motion of a quantum particle, a particle of small or large
mass. In particular, the states of atomic electrons are confined-wave states with wave functions
that are solutions to the Schrödinger equation.
*Q41.9 Answer (b). The reflected amplitude decreases as U decreases. The amplitude of the reflected
wave is proportional to the reflection coefficient, R, which is 1 − T , where T is the transmission
coefficient as given in equation 41.22. As U decreases, C decreases as predicted by equation
41.23, T increases, and R decreases.
*Q41.10 Answer (a). Because of the exponential tailing of the wave function within the barrier, the tunneling current is more sensitive to the width of the barrier than to its height.
Q41.11
Consider the Heisenberg uncertainty principle. It implies that electrons initially moving at the
same speed and accelerated by an electric field through the same distance need not all have
the same measured speed after being accelerated. Perhaps the philosopher could have said “it
is necessary for the very existence of science that the same conditions always produce the same
results within the uncertainty of the measurements.”
Q41.12 In quantum mechanics, particles are treated as wave functions, not classical particles. In classical
mechanics, the kinetic energy is never negative. That implies that E ≥ U . Treating the particle as
a wave, the Schrödinger equation predicts that there is a nonzero probability that a particle can
tunnel through a barrier—a region in which E < U .
*Q41.13 Answer (c). Other points see a wider potential-energy barrier and carry much less tunneling
current.
Quantum Mechanics
465
SOLUTIONS TO PROBLEMS
Section 41.1
P41.1
(a)
An Interpretation of Quantum Mechanics
ψ ( x ) = Ae (
i 5.00 ×1010 x
) = A cos 5 × 1010 x + Ai sin 5 × 1010 x = A cos ( kx ) + Ai sin ( kx ) goes
(
)
(
)
through a full cycle when x changes by l and when kx changes by 2π . Then k λ = 2π
2π
2π m
where k = 5.00 × 1010 m −1 =
. Then λ =
= 1.26 × 10 −10 m .
10
λ
5
00
10
×
.
(
)
h 6.626 × 10 −34 J ⋅ s
=
= 5.27 × 10 −24 kg ⋅ m s
λ
1.26 × 10 −10 m
(b)
p=
(c)
me = 9.11 × 10 −31 kg
−24
kg ⋅ m s )
1.52 × 10 −17 J
m 2u 2
p 2 ( 5.27 × 10
−17
1
52
10
K= e =
=
=
.
×
J
=
= 95.5 eV
2 me
2m
1.60 × 10 −19 J eV
( 2 × 9.11 × 10 −31 kg)
2
a
P41.2
Probability
P=
∫
ψ ( x) =
−a
P=
Section 41.2
P41.3
2
a
⎛ a ⎞ ⎛ 1⎞
−1 ⎛ x ⎞
∫− a π ( x 2 + a 2 ) dx = ⎝ π ⎠ ⎝ a ⎠ tan ⎝ a ⎠ − a
1
1 ⎡π ⎛ π ⎤
1
⎡ tan −1 1 − tan −1 ( −1) ⎤⎦ = ⎢ − − ⎞ ⎥ =
π⎣
π ⎣4 ⎝ 4⎠⎦
2
The Quantum Particle under Boundary Conditions
E1 = 2.00 eV = 3.20 × 10 −19 J
For the ground state,
P41.4
a
a
E1 =
h2
8 me L2
h
= 4.34 × 10 −10 m = 0.434 nm
8 me E1
(a)
L=
(b)
⎛ h2 ⎞ ⎛ h2 ⎞
= 6.00 eV
−
∆E = E2 − E1 = 4 ⎜
⎝ 8 me L2 ⎟⎠ ⎜⎝ 8 me L2 ⎟⎠
For an electron wave to “fit” into an infinitely deep potential well, an integral number
of half-wavelengths must equal the width of the well.
nλ
= 1.00 × 10 −9 m
2
(a)
(b)
so
λ=
2.00 × 10 −9 h
=
n
p
Since
(h2 λ 2 ) = h2
p2
n2
2
K=
=
2 = ( 0.377n ) eV
2 me
2 me
2 me ( 2 × 10 −9 )
For
K ≈ 6 eV
n=4
With
n = 4,
K = 6.03 eV
FIG. P41.4
466
P41.5
Chapter 41
(a)
We can draw a diagram that parallels our treatment of
standing mechanical waves. In each state, we measure the
distance d from one node to another (N to N), and base our
solution upon that:
dN to N =
Since
h
h
=
λ 2d
p=
2
−34
1 ⎡ ( 6.626 × 10 J ⋅ s ) ⎤
p2
h2
⎢
⎥
K=
=
=
2 me 8 me d 2 d 2 ⎢ 8 ( 9.11 × 10 −31 kg ) ⎥
⎣
⎦
Next,
Evaluating, K =
(b)
λ
h
and λ =
2
p
6.02 × 10 −38 J ⋅ m 2
d2
K=
3.77 × 10 −19 eV ⋅ m 2
d2
In state 1,
d = 1.00 × 10 −10 m
K1 = 37.7 eV
In state 2,
d = 5.00 × 10 −11 m
K 2 = 151 eV
In state 3,
d = 3.33 × 10 −11 m
K 3 = 339 eV
In state 4,
d = 2.50 × 10 −11 m
K 4 = 603 eV
FIG. P41.5
When the electron falls from state 2 to state 1, it puts out energy
E = 151 eV − 37.7 eV = 113 eV = hf =
hc
λ
into emitting a photon of wavelength
λ=
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
= 11.0 nm
E
(113 eV ) (1.60 × 10 −19 J eV )
The wavelengths of the other spectral lines we find similarly:
4→3
4→2
4 →1
3→2
3→1
2→1
E ( eV )
264
452
565
188
302
113
λ ( nm )
4.71
2.75
2.20
6.60
4.12
11.0
Transition
*P41.6
For the bead’s energy we have both (1/2)mu2 and h2n2/8mL2. Then
n=
2L
L
1
8 mL2 2 muL
mu 2 2 =
note that this expression can be thought of as
=
λ
d NN
2
h
h
Evaluating, n =
P41.7
2(0.005 kg)(10 −10 m) 0.2 m
= 9.56 × 1012
3.156 × 10 7 s (6.626 × 10 −34 J ⋅ s)
∆E =
hc ⎛ h 2 ⎞ 2 2
3h 2
=⎜
2
1
−
=
⎡
⎤
⎦ 8 m L2
λ ⎝ 8 me L2 ⎟⎠ ⎣
e
L=
3hλ
= 7.93 × 10 −10 m = 0.793 nm
8 me c
Quantum Mechanics
P41.8
P41.9
∆E =
hc ⎛ h 2 ⎞ 2 2
3h 2
=⎜
⎡ 2 − 1 ⎤⎦ =
2⎟ ⎣
λ ⎝ 8 me L ⎠
8 me L2
so
L=
467
3hλ
8 me c
The confined proton can be described in the same way as a
standing wave on a string. At level 1, the node-to-node distance
of the standing wave is 1.00 × 10 −14 m, so the wavelength is
twice this distance:
h
= 2.00 × 10 −14 m
p
The proton’s kinetic energy is
(6.626 × 10 −34 J ⋅ s )
p2
h2
1
mu 2 =
=
=
2
2
2 m 2 mλ 2 2 (1.67 × 10 −27 kg ) ( 2.00 × 10 −14 m )
2
K=
FIG. P41.9
3.29 × 10 −13 J
= 2.05 MeV
=
1.60 × 10 −19 J eV
In the first excited state, level 2, the node-to-node distance is half as long as in state 1. The
momentum is two times larger and the energy is four times larger: K = 8.22 MeV .
The proton has mass, has charge, moves slowly compared to light in a standing wave state, and
stays inside the nucleus. When it falls from level 2 to level 1, its energy change is
2.05 MeV − 8.22 MeV = −6.16 MeV
Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at the
speed of light, and that it has an energy of +6.16 MeV .
Its frequency is
f =
6
−19
J eV )
E ( 6.16 × 10 eV ) (1.60 × 10
=
= 1.49 × 10 21 Hz
−34
h
6.626 × 10 J ⋅ s
And its wavelength is
λ=
c 3.00 × 108 m s
=
= 2.02 × 10 −13 m
f
1.49 × 10 21 s −1
This is a gamma ray , according to the electromagnetic spectrum chart in Chapter 34.
P41.10
The ground state energy of a particle (mass m) in a 1-dimensional box of width L is E1 =
(a)
For a proton ( m = 1.67 × 10 −27 kg ) in a 0.200-nm wide box:
(b)
(6.626 × 10 J ⋅ s )
E =
= 8.22 × 10 J =
8 (1.67 × 10 kg ) ( 2.00 × 10 m )
For an electron ( m = 9.11 × 10 kg ) in the same size box:
(6.626 × 10 J ⋅ s )
E =
= 1.51 × 10 J =
8 ( 9.11 × 10 kg ) ( 2.00 × 10 m )
−34
2
−22
1
−27
−10
2
−10
2
−31
−34
2
−18
1
(c)
5.13 × 10 −3 eV
−31
9.41 eV
The electron has a much higher energy because it is much less massive.
h2
.
8 mL2
468
Chapter 41
⎛ h2 ⎞ 2
*P41.11 En = ⎜
n
⎝ 8 mL2 ⎟⎠
(6.626 × 10 J ⋅ s )
8 (1.67 × 10 kg ) ( 2.00 × 10
−34
E1 =
−27
E1 = 0.513 MeV
2
−14
m)
2
= 8.22 × 10 −14 J
E2 = 4 E1 = 2.05 MeV
E3 = 9 E1 = 4.62 MeV
Yes; the energy differences are ~1 MeV, which is a typical energy for a g -ray photon as radiated
by an atomic nucleus in an excited state.
P41.12
(a)
The energies of the confined electron are En =
jump from state 1 to state 4 is
h2
n 2 . Its energy gain in the quantum
8 me L2
h2
( 4 2 − 12 ) and this is the photon
8 me L2
⎛ 15hλ ⎞
h 2 15
hc
= hf =
energy:
. Then 8 me cL2 = 15hλ and L = ⎜
2
8 me L
λ
⎝ 8 me c ⎟⎠
(b)
Let λ ′ represent the wavelength of the photon emitted:
Then
*P41.13 (a)
(b)
(a)
hc
h2
h2
12h 2
42 −
22 =
=
.
2
2
λ ′ 8 me L
8 me L
8 me L2
From ∆x∆p ≥ h/2 with ∆x = L, the uncertainty in momentum must be at least ∆p ≈ h/2L .
Its energy is all kinetic, E = p2/2m = (∆p)2/2m ≈ h2/8mL2 = h2/(4p)28mL2.
Compared to the actual h2/8mL2, this estimate is too low by 4p2 ≈ 40 times. The actual wave
function does not have the particular (Gaussian) shape of a minimum-uncertainty
wave function. The result correctly displays the pattern of dependence of the energy
on the mass and on the length of the well.
x = ∫x
0
2 2 ⎛ 2π x ⎞
2
1 1
4π x ⎞
dx
sin
dx = ∫ x ⎛ − cos
⎝
⎠
⎝
L ⎠
L
L
L0 2 2
1 x2
x =
L 2
L
L
−
0
1 L2
L 16π 2
0.510 L
(b)
.
2
2
hc λ ′ h 15 (8 me L ) 5
=
= and λ ′ = 1.25λ .
2
2
λ hc
8 me L 12h
4
L
P41.14
12
Probability =
∫
0.490 L
⎡ 4π x sin 4π x + cos 4π x ⎤ = L
⎢⎣ L
2
L
L ⎥⎦ 0
L
2 2 ⎛ 2π x ⎞
1
1 L
4π x ⎤
dx = ⎡⎢ x −
sin
sin
⎝ L ⎠
L
L 4π
L ⎥⎦ 0.490 L
⎣L
Probability = 0.020 −
0.510 L
1
( sin 2.04π − sin 1.96π ) = 5.26 × 10 −5
4π
x
1
4π x ⎤
sin
= 3.99 × 10 −2
Probability ⎡⎢ −
L ⎥⎦ 0.240 L
⎣ L 4π
0.260 L
(c)
(d)
In the n = 2 graph in the text’s Figure 41.4(b), it is more probable to find the particle
L
3L
either near x =
or x =
than at the center, where the probability density is zero.
4
4
L
Nevertheless, the symmetry of the distribution means that the average position is .
2
Quantum Mechanics
P41.15
Normalization requires
L
∫
2
ψ dx = 1
or
L
0
2
nπ x ⎞
L
sin 2 ⎛
dx = A 2 ⎛ ⎞ = 1
⎝ L ⎠
⎝ 2⎠
The probability is
∫
2
ψ 1 dx =
0
2
L
nπ x ⎞
sin 2 ⎛
dx = 1
⎝ L ⎠
A=
or
L /3
*P41.16 (a)
2
0
all space
∫A
∫A
L /3
∫ sin
2
0
2
L
⎛ π x ⎞ dx = 1
⎝ L ⎠
L
L /3
⎛ 2π x ⎞ ⎤ dx
L ⎠ ⎥⎦
⎡
∫ ⎢⎣1 − cos ⎝
0
L /3
=
(b)
1⎡
2π x ⎞ ⎤
1 1
2π ⎞ 1 0.866
L
= −
sin ⎛
sin ⎛
x−
= −
= 0.196
⎝ L ⎠ ⎥⎦ 0
⎝ 3 ⎠ 3
2π
3 2π
L ⎢⎣
2π
Classically, the particle moves back and forth steadily, spending equal time intervals in
each third of the line. Then the classical probability is 0.333, significantly larger .
L /3
(c)
The probability is
∫
2
ψ 99 dx =
0
2
L
L /3
∫
0
sin 2 ⎛
⎝
99π x ⎞
1
dx =
L ⎠
L
L /3
⎡
⎛ 198π x ⎞ ⎤ dx
L ⎠ ⎥⎦
∫ ⎢⎣1 − cos ⎝
0
L /3
=
1⎡
198π x ⎞ ⎤
1
1
1
L
sin ⎛
= −
sin ( 66π ) = − 0 = 0.333
x−
⎝ L ⎠ ⎥⎦ 0
L ⎢⎣
198π
3 198π
3
in agreement with the classical model .
*P41.17 In 0 ≤ x ≤ L, the argument
2π x
of the sine function ranges from 0 to 2π . The probability
L
2
2π x ⎞
density ⎛ ⎞ sin 2 ⎛
reaches maxima at sin θ = 1 and sin θ = −1 . These points are at
⎝ L⎠
⎝ L ⎠
2π x π
2π x 3π
=
=
and
.
L
2
L
2
Therefore the most probable positions of the particle are at x =
L
3L
and x =
.
4
4
469
470
Chapter 41
*P41.18 (a)
Probability =
∫ψ
2
1
dx =
0
π x⎞
2
1 ⎡
2π x ⎞ ⎤
sin 2 ⎛
dx
dx = ∫ ⎢1 − cos ⎛
⎝ L ⎠
⎝ L ⎠ ⎥⎦
L ∫0
L 0⎣
=
1⎡
2π x ⎞ ⎤
1
2π ⎞
L
=
−
sin ⎛
sin ⎛
x−
⎥
⎢
⎝
⎝
⎠
2π
L⎣
L ⎦0
L 2π
L ⎠
(b)
Probability Curve for an Infinite
Potential Well
1.2
Probability
1
0.8
0.6
0.4
0.2
0
0.5
0
1
1.5
L
FIG. P41.18(b)
(c)
The wave function is zero for x < 0 and for x > L. The probability at l = 0 must be zero
because the particle is never found at x < 0 or exactly at x = 0. The probability at l = L must
be 1 for normalization: the particle is always found somewhere at x < L.
(d)
The probability of finding the particle between x = 0 and x = is
and x = L is
Thus,
∫ψ
2
1
1
.
3
dx =
0
∴
2
3
1
2π ⎞ 2
−
sin ⎛
= ,
⎝ L ⎠ 3
L 2π
This equation for
result being
2
, and between x = 3
or
u−
2
1
sin 2π u =
3
2π
can be solved by homing in on the solution with a calculator, the
L
= 0.585, or = 0.585 L
L
to three digits.
Quantum Mechanics
L3
P41.19
(a)
P=
The probability is
∫
L3
2
ψ dx =
0
∫
0
2 2 ⎛ π x⎞
2
dx =
sin
⎝
⎠
L
L
L
L3
⎛1
1
∫ ⎝ 2 − 2 cos
0
471
2π x ⎞
dx
L ⎠
⎛1
3⎞
x
1
2π x ⎞
1 1
2π
P=⎛ −
sin
=⎛ −
sin ⎞ = ⎜ −
= 0.196
⎝ L 2π
⎝ 3 2π
L ⎠0
3 ⎠ ⎝ 3 4π ⎟⎠
L3
(b)
(c)
Section 41.3
P41.20
L
.
2
Thus, the probability of finding the particle between
2L
x=
and x = L is the same 0.196. Therefore,
3
the probability of finding it in the range
L
2L
is P = 1.00 − 2 ( 0.196 ) = 0.609 .
≤x≤
3
3
The probability density is symmetric about x =
FIG. P41.19(b)
Classically, the electron moves back and forth with constant speed between the walls, and the
probability of finding the electron is the same for all points between the walls. Thus, the classical probability of finding the electron in any range equal to one-third of the available
1
space is Pclassical = . The result of part (a) is significantly smaller ,
3
because of the curvature of the graph of the probability density.
The Schrödinger Equation
ψ ( x ) = A cos kx + B sin kx
∂ψ
= − kA sin kx + kB cos kx
∂x
∂ 2ψ
= − k 2 A cos kx − k 2 B sin kx
∂x 2
−
2m
2 mE
( E − U )ψ = − 2 ( A cos kx + B sin kx )
Therefore the Schrödinger equation is satisfied if
∂ 2ψ ⎛ 2 m ⎞
= −
( E − U )ψ or
∂x 2 ⎝ 2 ⎠
2 mE
− k 2 ( A cos kx + B sin kx ) = ⎛ − 2 ⎞ ( A cos kx + B sin kx )
⎝ ⎠
This is true as an identity (functional equality) for all x if E =
P41.21
We have
ψ = Aei( kx −ω t )
so
∂ψ
= ikψ
∂x
and
We test by substituting into Schrödinger’s equation:
Since
k2 =
( 2π )2
λ2
=
( 2π p )2
h2
=
p2
2
and
2 k 2
.
2m
∂ 2ψ
= − k 2ψ .
∂x 2
∂ 2ψ
2m
= − k 2ψ = − 2 ( E − U ) ψ .
2
∂x
E −U =
p2
2m
Thus this equation balances.
472
P41.22
Chapter 41
(a)
Setting the total energy E equal to zero and
rearranging the Schrödinger equation to isolate
the potential energy function gives
⎛ 2 ⎞ 1 d 2ψ
U ( x) = ⎜
⎝ 2 m ⎟⎠ ψ dx 2
ψ ( x ) = Axe− x
If
2
L2
2
or
U ( x) =
and
(b)
P41.23
2
d 2ψ
e− x L
= ( 4 Ax 3 − 6 AxL2 ) 4
2
dx
L
2
2
2
d ψ (4 x − 6L )
=
ψ ( x)
dx 2
L4
Then
FIG. P41.22(b)
2 ⎛ 4 x 2
⎞
− 6⎟
⎠
2 mL2 ⎜⎝ L2
See the figure to the right.
Problem 41 in Chapter 16 helps students to understand how to draw conclusions from an identity.
(a)
⎛
x2 ⎞
ψ ( x ) = A ⎜1 − 2 ⎟
⎝
L ⎠
dψ
2 Ax
=− 2
dx
L
Schrödinger’s equation
d 2ψ
2m
= − 2 ( E − U )ψ
2
dx
becomes
−
2
2 2
2
⎛
2A
2m
x 2 ⎞ 2 m ( − x ) A (1 − x L )
+
=
−
−
EA
1
⎜⎝
L2
2
L2 ⎟⎠ 2
mL2 ( L2 − x 2 )
−
1
mE mEx 2 x 2
=
−
+ 2 2 − 4
L2
2
L
L
This will be true for all x if both
1 mE
= 2
L2
and
mE
1
− 4 =0
2 2
L
L
d 2ψ
2A
=− 2
dx 2
L
E=
both these conditions are satisfied for a particle of energy
1=
For normalization,
2
⎛
x2 ⎞
2x2 x4 ⎞
2 ⎛
2
∫− L A ⎜⎝ 1 − L2 ⎟⎠ dx = A −∫L ⎜⎝ 1 − L2 + L4 ⎟⎠ dx
L
(b)
2
L2 m
L
L
2x3
x5 ⎤
2
2
L
16 L ⎞
L
⎡
1 = A 2 ⎢ x − 2 + 4 ⎥ = A 2 ⎡⎢ L − L + + L − L + ⎤⎥ = A 2 ⎛
⎝
3L 5L ⎦− L
3
5
3
5⎦
15 ⎠
⎣
⎣
L3
(c)
P=
∫
−L 3
=
P=
L3
ψ 2 dx =
L3
⎛
15
2x2 x4 ⎞
15 ⎡
2x3 x5 ⎤
1
x
−
+
dx
=
−
+
16 L − L∫ 3 ⎜⎝
16 L ⎢⎣
3 L2 5 L5 ⎥⎦ − L 3
L2
L4 ⎟⎠
30 ⎡ L 2 L
L ⎤
−
+
16 L ⎢⎣ 3 81 1 215 ⎥⎦
47
= 0.580
81
A=
15
16 L
Quantum Mechanics
P41.24
(a)
ψ1 ( x) =
π x⎞
2
cos ⎛
;
⎝
L
L ⎠
P1 ( x ) = ψ 1 ( x ) =
π x⎞
2
cos 2 ⎛
⎝
L
L ⎠
ψ 2 ( x) =
2
2π x ⎞
sin ⎛
;
⎝
L
L ⎠
P2 ( x ) = ψ 2 ( x ) =
2 2 ⎛ 2π x ⎞
sin
⎝ L ⎠
L
ψ 3 ( x) =
2
3π x ⎞
cos ⎛
;
⎝
L
L ⎠
P3 ( x ) = ψ 3 ( x ) =
2
3π x ⎞
cos 2 ⎛
⎝
L
L ⎠
(b)
2
2
2
∞
∞
∞
473
∞
n=3
ψ
ψ
2
n=2
n=1
−
L
2
0
L
2
−
L
2
x
0
L
2
x
FIG. P41.24(b)
P41.25
(a)
With ψ ( x ) = A sin ( kx )
d2
ψ = − Ak 2 sin kx
dx 2
h 2 ( 4π 2 )
2 d 2ψ
2 k 2
p2
m 2u 2
1
A
kx
Then −
=+
sin =
ψ=
ψ=
ψ = mu 2ψ = Kψ
2
2
2
2 m dx
2m
2m
2m
2
4π ( λ ) ( 2 m )
d
A sin kx = Ak cos kx
dx
(b)
Section 41.4
and
2π x ⎞
= A sin kx , the proof given in part (a) applies again.
With ψ ( x ) = A sin ⎛
⎝ λ ⎠
A Particle in a Well of Finite Height
P41.26
FIG. P41.26
474
P41.27
Chapter 41
(a)
(b)
See figure to the right.
The wavelength of the transmitted wave
traveling to the left is the same as the original
wavelength, which equals 2L .
FIG. P41.27(a)
Section 41.5
P41.28
C=
Tunneling Through a Potential Energy Barrier
2 ( 9.11 × 10 −31 ) ( 5.00 − 4.50 ) (1.60 × 10 −19 ) kg ⋅ m s
1.055 × 10 −34 J ⋅ s
= 3.62 × 10 m
9
−1
T = e−2CL = exp ⎡⎣ −2 ( 3.62 × 10 9 m −1 ) ( 950 × 10 −12 m ) ⎤⎦ = exp ( −6.88 )
T = 1.03 × 10
P41.29
FIG. P41.28
−3
From problem 28, C = 3.62 × 10 9 m −1
We require 10 −6 = exp ⎡⎣ −2 ( 3.62 × 10 9 m −1 ) L ⎤⎦ .
Taking logarithms,
−13.816 = −2 ( 3.62 × 10 9 m −1 ) L
New L = 1.91 nm
∆L = 1.91 nm − 0.950 nm = 0.959 nm
*P41.30 T = e−2CL where C =
(a)
2CL =
(b)
2CL =
(c)
2CL =
(d)
2CL =
2 m (U − E )
2 2 ( 9.11 × 10 −31 ) ( 0.01 × 1.6 × 10 −19 )
1.055 × 10 −34
2 2 ( 9.11 × 10 −31 ) (1.6 × 10 −19 )
1.055 × 10 −34
(10 ) = 0.102
−10
(10 ) = 1.02
1.055 × 10 −34
2 2 (8 ) (1)
( 0.02 ) = 1.52 × 10 33
1.055 × 10 −34
T = e−1.02 = 0.359
−10
2 2 ( 6.65 × 10 −27 ) (10 6 × 1.6 × 10 −19 )
T = e−0.102 = 0.903
(10 ) = 0.875
−15
33
T = e−0.875 = 0.417
T = e−1.52×10 = e(ln10 )( −1.52×10
33
/ln 10 )
= 10 −6.59×10
32
Quantum Mechanics
P41.31
2CL =
P41.32
2 m (U − E )
T = e−2CL where C =
2 2 ( 9.11 × 10 −31 ) (8.00 × 10 −19 )
1.055 × 10 −34
( 2.000 × 10 ) = 4.58
−10
(a)
T = e−4.58 = 0.010 3 , a 1% chance of transmission.
(b)
R = 1 − T = 0.990 , a 99% chance of reflection.
FIG. P41.31
The original tunneling probability is T = e−2CL where
C=
( 2m (U − E ))1 2
2π ( 2 × 9.11 × 10 −31 kg ( 20 − 12 ) 1.6 × 10 −19 J )
12
=
6.626 × 10 −34 J ⋅ s
The photon energy is hf =
= 1.448 1 × 1010 m −1
hc 1 240 eV ⋅ nm
=
= 2.27 eV, to make the electron’s new kinetic
λ
546 nm
energy 12 + 2.27 = 14.27 eV and its decay coefficient inside the barrier
2π ( 2 × 9.11 × 10 −31 kg ( 20 − 14.27 ) 1.6 × 10 −19 J )
12
C′ =
6.626 × 10 −34 J ⋅ s
= 1.225 5 × 1010 m −1
Now the factor of increase in transmission probability is
−9
10
−1
e−2C ′L
= e2 L(C −C ′ ) = e2×10 m × 0.223×10 m = e4.45 = 85.9
−2 CL
e
Section 41.6
P41.33
Applications of Tunneling
− CL
With the wave function proportional to e , the transmission coefficient and the tunneling
current are proportional to ψ , to e−2CL .
2
Then,
P41.34
I ( 0.500 nm ) e−2(10.0 nm )( 0.500 nm )
=
= e20.0( 0.015) = 1.35
I ( 0.515 nm ) e−2(10.0 nm )( 0.515 nm )
With transmission coefficient e−2CL , the fractional change in transmission is
e−2(10.0 nm )L − e−2(10.0 nm )( L + 0.002 00 nm )
= 1 − e−20.0( 0.002 00 ) = 0.0392 = 3.92%
e−2(10.0 nm )L
475
476
Chapter 41
The Simple Harmonic Oscillator
Section 41.7
P41.35
ψ = Be−( mω 2 )x so
2
dψ
mω ⎞
d 2ψ ⎛ mω ⎞ 2
mω ⎞
= −⎛
=
xψ and
x ψ + ⎛−
ψ
2
⎝
⎠
⎝
⎠
⎝
⎠
dx
dx
2
Substituting into the Schrödinger equation gives
⎛ mω ⎞ x 2ψ + ⎛ − mω ⎞ ψ = − ⎛ 2 mE ⎞ ψ + ⎛ mω ⎞ x 2ψ
⎝ ⎠
⎝ ⎠
⎝ 2 ⎠
⎝ ⎠
2
2
which is satisfied provided that E =
P41.36
ω
.
2
Problem 41 in Chapter 16 helps students to understand how to draw conclusions from an identity.
ψ = Axe− bx so
2
2
dψ
= Ae− bx − 2bx 2 Ae− bx
dx
and
2
2
2
d 2ψ
= −2bxAe− bx − 4bxAe− bx + 4b 2 x 3 e− bx = −6bψ + 4b 2 x 2ψ
2
dx
2
2 mE ⎞
mω ⎞ 2
−6bψ + 4b 2 x 2ψ = − ⎛
ψ +⎛
xψ
⎝ ⎠
⎝ ⎠
2
Substituting into the Schrödinger equation,
For this to be true as an identity, it must be true for all values of x.
So we must have both
P41.37
(a)
Therefore
(b)
and
(c)
The wave function is that of the
−6b = −
b=
E=
2
2 mE
⎛ mω ⎞
2
and
4
b
=
⎝ ⎠
2
mω
2
3b 2
3
ω
=
m
2
first excited state .
The longest wavelength corresponds to minimum photon energy, which must be equal to the
spacing between energy levels of the oscillator:
⎛ 9.11 × 10 −31 kg ⎞
m
hc
k
= ω = so λ = 2π c
= 2π ( 3.00 × 108 m s ) ⎜
⎝ 8.99 N m ⎟⎠
k
λ
m
12
= 600 nm
Quantum Mechanics
P41.38
(a)
With ψ = Be−( mω 2 )x , the normalization condition
2
∫
2
ψ dx = 1
all x
∞
becomes 1 =
2 −2( mω 2 ) x 2
∫B e
−∞
∞
dx = 2 B 2 ∫ e−( mω )x dx = 2 B 2
2
0
1
π
2 mω where Table B.6 in Appendix B was used to evaluate the integral.
Thus, 1 = B
(b)
2
⎛ mω ⎞
π
and B = ⎜
⎝ π ⎟⎠
mω
14
.
For small d, the probability of finding the particle in the range −
δ 2
⎛ mω ⎞
ψ dx = δ ψ ( 0 ) = δ B e = δ ⎜
∫
⎝ π ⎟⎠
−δ 2
2
P41.39
(a)
δ
δ
< x < is
2
2
12
2 −0
2
For the center of mass to be fixed, m1u1 + m2 u2 = 0. Then
u = u1 + u2 = u1 +
Similarly, u =
and u2 =
m1
m + m1
u1 = 2
u1
m2
m2
and
u1 =
m2 u
m1 + m2
m2
u2 + u2
m1
m1u
. Then
m1 + m2
1
1
1
1 m1 m22 u 2
1 m2 m12 u 2
1 2
kx
m1u12 + m2 u22 + kx 2 =
2 +
2 +
2
2
2
2 ( m1 + m2 )
2 ( m1 + m2 )
2
=
(b)
1 m1 m2 ( m1 + m2 ) 2 1 2 1 2 1 2
u + kx = µu + kx
2
2 ( m1 + m2 )
2
2
2
d ⎛1
1
µ u 2 + kx 2 ⎞ = 0 because energy is constant
⎠
dx ⎝ 2
2
0=
1
du 1
dx du
du
µ 2u
+ k 2x = µ
+ kx = µ
+ kx
2
dx 2
dt dx
dt
kx
. This is the condition for simple harmonic motion, that the
µ
acceleration of the equivalent particle be a negative constant times the excursion from
Then µ a = − kx , a = −
equilibrium. By identification with a = −ω 2 x , ω =
k
= 2π f and
µ
f =
1
2π
k
.
µ
477
478
P41.40
Chapter 41
(a)
With x = 0 and px = 0, the average value of x 2 is ( ∆x ) and the average value of px2
2
is ( ∆px ) . Then ∆x ≥
2
E≥
(b)
requires
2 ∆px
px2 k 2
px2 k 2
+
=
+
2 m 2 4 px2
2 m 8 px2
1
1
dE
k2
=0=
+
( −1) 4
2
2m
8
dpx
px
To minimize this as a function of px2 , we require
Then
k2
1
=
8 px4 2 m
and
E≥
Emin =
2
⎛ 2 mk 2 ⎞
px2 = ⎜
⎝ 8 ⎟⎠
so
mk
k2 2
+
=
2 ( 2 m ) 8 mk 4
k +
m 4
12
=
mk
2
k
m
k
ω
=
m
2
Additional Problems
P41.41
Suppose the marble has mass 20 g. Suppose the wall of the box is 12 cm high and 2 mm thick.
While it is inside the wall,
U = mgy = ( 0.02 kg ) ( 9.8 m s 2 ) ( 0.12 m ) = 0.023 5 J
and
E=K=
1
1
2
mu 2 = ( 0.02 kg ) ( 0.8 m s ) = 0.006 4 J
2
2
2 m (U − E )
=
Then C =
2 ( 0.02 kg ) ( 0.017 1 J )
1.055 × 10 −34 J ⋅ s
= 2.5 × 10 32 m −1
and the transmission coefficient is
e−2CL = e
P41.42
(
)(
−2 2.5×1032 2 ×10 −3
) = e−10 ×1029 = e−2.30( 4.3×1029 ) = 10 −4.3×1029 = ~ 10 −1030
(a)
λ = 2 L = 2.00 × 10 −10 m
(b)
p=
h 6.626 × 10 −34 J ⋅ s
=
= 3.31 × 10 −24 kg ⋅ m s
λ
2.00 × 10 −10 m
(c)
E=
p2
= 0.172 eV
2m
Quantum Mechanics
P41.43
(a)
See the figure.
(b)
See the figure.
FIG. P41.43(a)
FIG. P41.43(b)
(c)
y is continuous and ψ → 0 as x → ±∞. The function can be normalized. It describes a
particle bound near x = 0.
(d)
Since y is symmetric,
∞
∫
−∞
∞
479
ψ dx = 2 ∫ ψ dx = 1
2
2
0
∞
or
⎛ 2 A2 ⎞ − ∞ 0
2 A 2 ∫ e−2α x dx = ⎜
(e − e ) = 1
⎝ −2α ⎟⎠
0
This gives A = α .
(e)
P( −1 2α )→(1 2α ) = 2
1 2α
2α ⎞ − 2α 2α
e−2α x dx = ⎛
− 1) = (1 − e−1 ) = 0.632
(e
⎝
⎠
α
−
2
x=0
( a) ∫
2
*P41.44 If we had n = 0 for a quantum particle in a box, its momentum would be zero. The uncertainty in
its momentum would be zero. The uncertainty in its position would not be infinite, but just equal
to the width of the box. Then the uncertainty product would be zero, to violate the uncertainty
principle. The contradiction shows that the quantum number cannot be zero. In its ground state
the particle has some nonzero zero-point energy.
*P41.45 (a)
With ground state energy 0.3 eV, the energy in the n = 2 state is 22 × 0.3 eV = 1.2 eV. The
energy in state 3 is 9 × 0.3 eV = 2.7 eV. The energy in state 4 is 16 × 0.3 eV = 4.8 eV.
For the transition from the n = 3 level to the n = 1 level, the electron loses energy
(2.7 – 0.3) eV = 2.4 eV. The photon carries off this energy and has wavelength
hc/E = 1240 eV⋅nm/2.4 eV = 517 nm .
(b)
For the transition from level 2 to level 1, the photon energy is 0.9 eV and the photon
wavelength is l = hc/E = 1240 eV ⋅ nm/0.9 eV = 1.38 mm . This photon, with wavelength
greater than 700 nm, is infrared .
For level 4 to 1, E = 4.5 eV and l = 276 nm ultraviolet .
For 3 to 2, E = 1.5 eV and l = 827 nm infrared .
For 4 to 2, E = 3.6 eV and l = 344 nm near ultraviolet .
For 4 to 3, E = 2.1 eV and l = 590 nm yellow-orange visible .
480
P41.46
Chapter 41
(a)
Use Schrödinger’s equation
∂ 2ψ
2m
= − 2 ( E − U )ψ
2
∂x
with solutions
ψ 1 = Aeik1x + Be− ik1x
[region I ]
ψ 2 = Ceik2 x
[region II ]
FIG. P41.46(a)
Where
k1 =
2 mE
and
k2 =
2m ( E − U )
Then, matching functions and derivatives at x = 0
(ψ 1 )0 = (ψ 2 )0
gives
A+B=C
dψ
dψ
and ⎛ 1 ⎞ = ⎛ 2 ⎞
⎝ dx ⎠ 0 ⎝ dx ⎠ 0
gives
k1 ( A − B ) = k2 C
Then
B=
1 − k2 k1
A
1 + k2 k1
and
C=
2
A
1 + k2 k1
Incident wave Aeikx reflects Be− ikx , with probability
B 2 (1 − k2 k1 )
R= 2 =
A
(1 + k2 k1 )2
2
=
(b)
( k1 − k2 )2
( k1 + k2 )2
With
E = 7.00 eV
and
U = 5.00 eV
k2
=
k1
E −U
=
E
2.00
= 0.535
7.00
(1 − 0.535)2
= 0.092 0
(1 + 0.535)2
The reflection probability is
R=
The probability of transmission is
T = 1 − R = 0.908
Quantum Mechanics
P41.47
R=
481
( k1 − k2 )2 = (1 − k2 k1 )2
( k1 + k2 )2 (1 + k2 k1 )2
2 k 2
= E − U for constant U
2m
2 k12
= E since U = 0
2m
(1)
k
= E −U
2m
(2)
2
P41.48
FIG. P41.47
2
2
Dividing (2) by (1),
k22
U
k
1
1 1
= 1 − = 1 − = so 2 =
2
k1
E
k1
2 2
2
and therefore,
(1 − 1
R=
(1 + 1
) =(
2)
(
2
2
2
)
2 + 1)
2 −1
2
2
= 0.029 4
(a)
The wave functions and probability densities are the same as those shown in the two lower
curves in Figure 41.4 of the textbook.
(b)
P1 =
2 ⎞
πx ⎞
2
dx
ψ 1 dx = ⎛
sin 2 ⎛
∫
∫
⎝
⎠
⎝
1.00 nm ⎠
1.00 nm 0.150
0.150 nm
0.350 nm
0.350
0.350 nm
2π x ⎞ ⎤
⎡ x 1.00 nm
= ( 2.00 nm ) ⎢ −
sin ⎛
⎝
4π
1.00 nm ⎠ ⎥⎦ 0.150 nm
⎣2
x
1
In the above result we used ∫ sin 2 axdx = ⎛ ⎞ − ⎛ ⎞ sin ( 2ax ).
⎝ 2 ⎠ ⎝ 4a ⎠
0.350 nm
1.00 nm
2π x ⎞ ⎤
⎡
sin ⎛
Therefore, P1 = (1.00 nm ) ⎢ x −
⎝ 1.00 nm ⎠ ⎥⎦ 0.150 nm
2π
⎣
{
P1 = (1.00 nm ) 0.350 nm − 0.150 nm −
0.350
2
⎡ x 1.00 ⎛ 4π x ⎞ ⎤
2 ⎛ 2π x ⎞
∫ sin ⎝ 1.00 ⎠ dx = 2.00 ⎢⎣ 2 − 8π sin ⎝ 1.00 ⎠ ⎥⎦0.150
1.00 0.150
0.350
(c)
P2 =
}
1.00 nm
[sin ( 0.700π ) − sin ( 0.300π )] = 0.200
2π
0.350
1.00 ⎛ 4π x ⎞ ⎤
⎡
P2 = 1.00 ⎢ x −
sin
⎝ 1.00 ⎠ ⎥⎦ 0.150
4π
⎣
{
= 1.00 ( 0.350 − 0.150 ) −
}
1.00
[sin (1.40π ) − sin ( 0.600π )]
4π
= 0.351
(d)
Using En =
n2h2
, we find that E1 = 0.377 eV and E2 = 1.51 eV .
8 mL2
482
P41.49
P41.50
Chapter 41
(a)
f =
⎛ 1.60 × 10 −19 J ⎞
E
(1.80 eV )
= 4.34 × 1014 Hz
=
−34
h ( 6.626 × 10 J ⋅ s ) ⎜⎝ 1.00 eV ⎟⎠
(b)
λ=
c 3.00 × 108 m s
=
= 6.91 × 10 −7 m = 691 nm
f
4.34 × 1014 Hz
(c)
∆E ∆t ≥
(a)
Taking Lx = Ly = L , we see that the expression for E becomes
6.626 × 10 −34 J ⋅ s
h
=
=
= 2.64 × 10 −29 J = 1.65 × 10 −10 eV
so ∆E ≥
2 ∆t 4π ( ∆t ) 4π ( 2.00 × 10 −6 s )
2
E=
h2
nx2 + ny2
8 me L2
(
)
For a normalizable wave function describing a particle, neither nx nor ny can be zero. The
ground state, corresponding to nx = ny = 1, has an energy of
E1,1 =
h2
h2
12 + 12 ) =
2 (
8 me L
4 me L2
The first excited state, corresponding to either nx = 2, ny = 1 or nx = 1, ny = 2, has an
energy
E2,1 = E1,2 =
5h 2
h2
22 + 12 ) =
2 (
8 me L
8 me L2
The second excited state, corresponding to nx = 2, ny = 2, has an energy of
E 2 ,2 =
h2
h2
22 + 22 ) =
2 (
8 me L
me L2
Finally, the third excited state, corresponding to either nx = 1, ny = 3 or nx = 3, nx = 1, has
an energy
E1, 3 = E3,1 =
(b)
5h 2
h2
2
2
1
3
+
=
(
)
4 me L2
8 me L2
The energy difference between the
second excited state and the ground
state is given by
∆E = E2, 2 − E1,1 =
3h 2
=
4 me L2
h2
h2
−
2
me L 4 me L2
energy
E1, 3 , E3, 1
E2, 2
h2
me L2
E1, 2 , E2, 1
E1, 1
Energy level diagram
FIG. P41.50(b)
0
Quantum Mechanics
∞
P41.51
∫x
x2 =
2
ψ dx
2
−∞
For a one-dimensional box of width L, ψ n =
2
nπ x ⎞
.
sin ⎛
⎝
L
L ⎠
L2
L2
2 2 2 ⎛ nπ x ⎞
x
sin
dx
=
−
⎝ L ⎠
L ∫0
3 2 n 2π 2
L
Thus, x 2 =
∞
P41.52
(a)
∫ψ
2
(from integral tables).
dx = 1 becomes
−∞
L 4
2π x ⎞
L π
L ⎡ π x 1 ⎛ 4π x ⎞ ⎤
= A 2 ⎛ ⎞ ⎡⎢ ⎤⎥ = 1
cos 2 ⎛
+ sin
dx = A 2 ⎛ ⎞ ⎢
∫
⎥
⎝
⎝
⎠
⎝
⎠
⎝
⎠
2
π
4
L
2
π⎠ ⎣2⎦
L
L
⎦− L 4
⎣
−L 4
L 4
A2
or A 2 =
(b)
4
2
and A =
.
L
L
The probability of finding the particle between 0 and
L8
∫
L8
2
ψ dx = A 2
0
P41.53
∫ cos
2
0
L
is
8
⎛ 2π x ⎞ dx = 1 + 1 = 0.409
⎝ L ⎠
4 2π
For a particle with wave function
ψ ( x) =
2 −x a
e
a
for x > 0
for x < 0
and 0
(a)
ψ ( x ) = 0, x < 0
(b)
Prob ( x < 0 ) =
2
0
∫
−∞
Normalization
2 −2 x a
e , x > 0 as shown
a
0
ψ ( x ) dx = ∫ ( 0 ) dx = 0
2
−∞
∞
(c)
ψ 2 ( x) =
and
∫
ψ ( x ) dx =
2
−∞
0
0
∫
−∞
∞
ψ dx + ∫ ψ dx = 1
2
2
0
∞
⎛ 2 ⎞ −2 x a
−2 x a ∞
−∞
∫−∞0dx + ∫0 ⎝ a ⎠ e dx = 0 − e 0 = − ( e − 1) = 1
Prob ( 0 < x < a ) =
a
a
2
⎛ 2 ⎞ −2 x a
∫0 ψ dx = ∫0 ⎝ a ⎠ e dx
a
= − e−2 x a 0 = 1 − e−2 = 0.865
FIG. P41.53
483
484
P41.54
Chapter 41
(a)
The requirement that
E=
( pc )2 + ( mc 2 )
2
h nh
nλ
= L so p = =
is still valid.
2
λ 2L
2
2
nhc ⎞
⇒ En = ⎛
+ ( mc 2 )
⎝ 2L ⎠
2
K n = En − mc 2 =
(b)
⎛ nhc ⎞ + mc 2 2 − mc 2
)
⎝ 2L ⎠ (
Taking L = 1.00 × 10 −12 m, m = 9.11 × 10 −31 kg, and n = 1, we find K1 = 4.69 × 10 −14 J .
(6.626 × 10 −34 J ⋅ s )
h2
−14
Nonrelativistic, E1 =
J.
=
2 = 6.02 × 10
8 mL2 8 ( 9.11 × 10 −31 kg ) (1.00 × 10 −12 m )
2
Comparing this to K1 , we see that this value is too large by 28.6% .
P41.55
2
7 ke e 2
e2
1 1 ⎛
1⎞
⎤ ( − 7 3) e
⎡
)
=
1
=
−
−
+
−
+
−
+
+
−
1
1
(
⎥⎦ 4π ∈ d
4π ∈0 d ⎢⎣
3d
2 3 ⎝
2⎠
0
(a)
U=
(b)
From Equation 41.14, K = 2 E1 =
(c)
E = U + K and
2h 2
h2
=
.
2
36 me d 2
8 me ( 9 d )
7 ke e 2
h2
−
=0
3d 2 18 me d 3
dE
= 0 for a minimum:
dd
6.626 × 10 −34 )
(
3h 2
h2
d=
=
=
( 7 ) (18 ke e2 me ) 42 me ke e2 ( 42 ) ( 9.11 × 10 −31 ) (8.99 × 10 9 ) (1.60 × 10 −19 C )2
2
= 0.049 9 nm
(d)
Nm
Since the lithium spacing is a, where Na 3 = V , and the density is
, where m is the
V
mass of one atom, we get:
Vm ⎞
a=⎛
⎝ Nm ⎠
13
⎛ m ⎞
=⎜
⎝ density ⎟⎠
13
⎛ 1.66 × 10 −27 kg × 7 ⎞
=⎜
⎟⎠
⎝
530 kg
13
m = 2.80 × 10 −10 m = 0.280 nm
The lithium interatomic spacing of 280 pm is 5.62 times larger than the answer to (c). Thus
it is of the same order of magnitude as the interatomic spacing 2d here.
Quantum Mechanics
P41.56
(a)
ψ = Bxe−( mω 2 )x
485
2
2
2
2
mω ⎞ 2 −( mω 2 )x 2
dψ
mω ⎞
= Be−( mω 2 )x + Bx ⎛ −
x e
2 xe−( mω 2 )x = Be−( mω 2 )x − B ⎛
⎝ 2 ⎠
⎝ ⎠
dx
2
mω ⎞
mω ⎞ 2 ⎛ mω ⎞ −( mω 2 )x 2
d 2ψ
mω ⎞ −( mω 2 )x 2
= Bx ⎛ −
− B⎛
xe
2 xe−( mω 2 )x − B ⎛
x −
xe
2
⎝
⎠
⎝
⎠
⎝
dx
⎠ ⎝ ⎠
mω ⎞ 3 −( mω 2 )x 2
d 2ψ
mω ⎞ −( mω 2 )x 2
= −3 B ⎛
xe
+ B⎛
xe
⎝ ⎠
⎝ ⎠
dx 2
2
Substituting into the Schrödinger equation, we have
2
2
2 mE
mω ⎞ 2
mω ⎞ −( mω 2 )x 2
mω ⎞ 3 −( mω 2 )x 2
x Bxe−( mω 2 )x
−3 B ⎛
xe
+ B⎛
xe
= − 2 Bxe−( mω 2 )x + ⎛
⎝ ⎠
⎝ ⎠
⎝ ⎠
2
2
2E
3ω
; it is true if E =
.
2
This is true if −3ω = −
(b)
We never find the particle at x = 0 because ψ = 0 there.
(c)
y is maximized if
(d)
We require
∞
∫ψ
2
dψ
mω ⎞
= 0 = 1 − x2 ⎛
, which is true at x = ±
.
⎝ ⎠
dx
mω
dx = 1:
−∞
∞
1=
2 2 −( mω ) x
dx = 2 B 2 ∫ x 2 e−( mω )x dx = 2 B 2
∫B x e
2
2
−∞
Then B =
(e)
(f)
21 2 ⎛ mω ⎞
π1 4 ⎝ ⎠
34
⎛ 4 m 3ω 3 ⎞
= ⎜
⎝ π 3 ⎟⎠
1
4
π
B2 π 1 2 3 2
3 =
32
2 ( mω )
( mω )
14
.
1
1
4 ⎞
, the potential energy is mω 2 x 2 = mω 2 ⎛
= 2ω . This is larger than
⎝
mω
mω ⎠
2
2
3ω
the total energy
, so there is zero classical probability of finding the particle here.
2
At x = 2
(
−( mω 2 ) x
Probability = ψ dx = Bxe
2
Probability = δ
2 ⎛ mω ⎞
π1 2 ⎝ ⎠
32
2
) δ = δB x e
2
2
2 −( mω ) x 2
⎛ 4 ⎞ e−( mω )4( mω ) = 8δ ⎛ mω ⎞
⎝ mω ⎠
⎝ π ⎠
12
e−4
486
Chapter 41
π x⎞
π x ⎞ ⎛ 2π x ⎞ ⎤
2π x ⎞
⎡
+ 16 sin 2 ⎛
+ 8 sin ⎛
dx = 1
A 2 ∫ ⎢sin 2 ⎛
sin
⎝
⎠
⎝
⎠
⎝
⎝ L ⎠ ⎥⎦
L
L
L ⎠
0 ⎣
L
P41.57
(a)
∫ψ
L
2
dx = 1 :
0
L
⎡ L
π x ⎞ ⎛ 2π x ⎞ ⎤
L
sin
A 2 ⎢⎛ ⎞ + 16 ⎛ ⎞ + 8 ∫ sin ⎛
dx = 1
⎝
⎝ L ⎠ ⎥
⎝ 2⎠
⎝
⎠
2
L ⎠
0
⎣
⎦
L
⎡ 17 L 16 L 3 ⎛ π x ⎞
⎡ 17 L
π x⎞
π x⎞ ⎤
+ 16 ∫ sin 2 ⎛
A2 ⎢
cos ⎛
dx ⎥ = A 2 ⎢
+
sin
⎝ L ⎠
⎝ L ⎠
⎝ L ⎠
3π
0
⎣ 2
⎦
⎣ 2
A2 =
∫
⎤
⎥ =1
x=0 ⎦
2
2
, so the normalization constant is A =
.
17 L
17 L
a
(b)
x= L
a
⎡
∫ ⎢⎣ A
2
ψ dx = 1:
−a
2
−a
π x⎞
π x⎞
π x⎞ ⎛ π x⎞ ⎤
2
cos 2 ⎛
dx = 1
sin
+ B sin 2 ⎛
+ 2 A B cos ⎛
⎝ 2a ⎠
⎝ a ⎠ ⎥⎦
⎝ a ⎠
⎝ 2a ⎠
2
2
The first two terms are A a and B a . The third term is:
⎛ π x⎞ ⎡
a
⎛ π x⎞
⎛ π x⎞ ⎤
∫ cos ⎝ 2a ⎠ ⎢⎣ 2 sin ⎝ 2a ⎠ cos ⎝ 2a ⎠ ⎥⎦ dx = 4 A
2A B
−a
a
B
∫ cos
−a
2
⎛ π x ⎞ sin ⎛ π x ⎞ dx
⎝ 2a ⎠
⎝ 2a ⎠
a
=
(
so that a A + B
2
∞
P41.58
x
(b)
x1=
(c)
x
0
⎛ a⎞
∫−∞ x ⎝ π ⎠
=
(a)
∞
A + B =
2
2
1
.
a
2
e− ax dx = 0 , since the integrand is an odd function of x.
⎛ 4a3 ⎞
∫−∞ x ⎜⎝ π ⎟⎠
=
) = 1, giving
12
∞
01
2
8a A B
π x⎞
cos3 ⎛
=0
⎝
3π
2a ⎠ − a
1
∫ x 2 (ψ
−∞
12
2
x 2 e− ax dx = 0 , since the integrand is an odd function of x.
+ ψ 1 ) dx =
2
0
1
x
2
∞
0
+
1
x + ∫ xψ 0 ( x ) ψ 1 ( x ) dx
2 1 −∞
The first two terms are zero, from (a) and (b). Thus:
∞
x
01
a
= ∫ x⎛ ⎞
⎝
π⎠
−∞
14
⎛ 2a 2 ⎞
= 2⎜
⎝ π ⎟⎠
12
=
1
2a
e
− ax 2 2
⎛ 4a3 ⎞
⎜⎝ π ⎟⎠
14
xe
− ax 2 2
⎛ 2a 2 ⎞
dx = 2 ⎜
⎝ π ⎟⎠
1⎛ π ⎞
, from Table B.6
4 ⎝ a3 ⎠
12
12 ∞
∫x e
0
2 − ax 2
dx
Quantum Mechanics
P41.59
2
With one slit open
P1 = ψ 1 or P2 = ψ 2
With both slits open,
P = ψ1 + ψ 2
At a maximum, the wave functions are in phase
Pmax = ( ψ 1 + ψ 2
)
At a minimum, the wave functions are out of phase
Pmin = ( ψ 1 − ψ 2
)
2
Now
P1 ψ 1
=
2 = 25.0 , so
P2 ψ 2
( ψ1 + ψ 2
P
and max =
Pmin ( ψ 1 − ψ 2
)
)
2
2
(5.00 ψ
=
(5.00 ψ
487
2
2
2
2
ψ1
= 5.00
ψ2
2
+ ψ2
2
− ψ2
)
)
2
2
=
( 6.00 )2 36.0
=
= 2.25
( 4.00 )2 16.0
ANSWERS TO EVEN PROBLEMS
P41.2
1
2
P41.4
(a) 4
P41.6
9.56 × 1012
P41.8
⎛ 3hλ ⎞
⎜⎝ 8 m c ⎟⎠
e
P41.10
(b) 6.03 eV
12
(a) 5.13 meV (b) 9.41 eV (c) The much smaller mass of the electron requires it to have much
more energy to have the same momentum.
12
P41.12
P41.14
P41.16
P41.18
⎛ 15hλ ⎞
(a) ⎜
(b) 1.25λ
⎝ 8 me c ⎟⎠
L
(a)
(b) 5.26 × 10 −5 (c) 3.99 × 10 −2
2
(d) See the solution.
(a) 0.196 (b) The classical probability is 0.333, significantly larger.
and quantum models.
(c) 0.333 for both classical
1
2π ⎞
(b) See the solution. (c) The wave function is zero for x < 0 and for x > L.
−
sin ⎛
⎝
L 2π
L ⎠
The probability at l = 0 must be zero because the particle is never found at x < 0 or exactly at x = 0.
The probability at l = L must be 1 for normalization. This statement means that the particle is always
found somewhere at x < L. (d) l = 0.585L
(a)
P41.20
See the solution;
P41.22
(a)
2 k 2
2m
2 ⎛ 4 x 2
⎞
− 6⎟
⎠
2 mL2 ⎜⎝ L2
(b) See the solution.
488
P41.24
Chapter 41
(a) ψ 1 ( x ) =
π x⎞
2
cos ⎛
⎝ L ⎠
L
P1 ( x ) =
P2 ( x ) =
2 2 ⎛ 2π x ⎞
sin
⎝ L ⎠
L
ψ 3 ( x) =
2
π x⎞
cos 2 ⎛
⎝ L ⎠
L
2
3π x ⎞
cos ⎛
⎝
L
L ⎠
ψ 2 ( x) =
P3 ( x ) =
2
2π x ⎞
sin ⎛
⎝ L ⎠
L
2
3π x ⎞
cos 2 ⎛
⎝
L
L ⎠
(b) See the solution.
P41.26
See the solution.
P41.28
1.03 × 10 −3
P41.30
(a) 0.903
P41.32
85.9
P41.34
3.92%
P41.36
(a) See the solution.
P41.38
⎛ mω ⎞
(a) B = ⎜
⎝ π ⎟⎠
P41.40
See the solution.
P41.42
(a) 2.00 × 10 −10 m
P41.44
See the solution.
P41.46
(a) See the solution.
(b) 0.092 0 , 0.908
P41.48
(a) See the solution.
(b) 0.200
(b) 0.359
P41.52
b=
mω
2
⎛ mω ⎞
(b) δ ⎜
⎝ π ⎟⎠
(b) E =
3
ω
2
2
(c) first excited state
(c) 0.351
(c) 0.172 eV
(d) 0.377 eV, 1.51 eV
2
h
5h
h
5h
,
,
,
2
2
2
4 me L 8 me L me L 4 me L2
2
(a)
(b) 0.409
L
(a)
32
12
(b) 3.31 × 10 −24 kg ⋅ m s
2
2
P41.50
14
(d) 10 −6.59×10
(c) 0.417
(b) See the solution,
3h 2
4 me L2
2
P41.54
P41.56
P41.58
nhc ⎞
(a) ⎛
+ m 2 c 4 − mc 2 (b) 46.9 fJ; 28.6%
⎝ 2L ⎠
14
⎛ 4 m 3ω 3 ⎞
3ω
(a)
(b) x = 0 (c) ±
(d) ⎜
⎝ π 3 ⎟⎠
2
mω
(a) 0
(b) 0
(c) ( 2a )
−1 2
mω ⎞
(e) 0 (f ) 8δ ⎛
⎝ π ⎠
12
e−4
42
Atomic Physics
Note:
In chapters 39, 40, and 41 we used u to represent the speed of a particle with mass. In this chapter 42
and the remaining chapters we go back to using v for the symbol for speed.
CHAPTER OUTLINE
42.1
42.2
42.3
Atomic Spectra of Gases
Early Models of the Atom
Bohr’s Model of the
Hydrogen Atom
42.4 The Quantum Model of the
Hydrogen Atom
42.5 The Wave Functions of
Hydrogen
42.6 Physical Interpretation of the
Quantum Numbers
42.7 The Exclusion Principle and
the Periodic Table
42.8 More on Atomic Spectra:
Visible and X-ray
42.9 Spontaneous and Stimulated
Transitions
42.10 Lasers
ANSWERS TO QUESTIONS
Q42.1
If an electron moved like a hockey puck, it could have
any arbitrary frequency of revolution around an atomic
nucleus. If it behaved like a charge in a radio antenna,
it would radiate light with frequency equal to its own
frequency of oscillation. Thus, the electron in hydrogen
atoms would emit a continuous spectrum, electromagnetic
waves of all frequencies smeared together.
*Q42.2 (a)
(b)
Yes, provided that the energy of the photon is
precisely enough to put the electron into one of the
allowed energy states. Strangely—more precisely
non-classically—enough, if the energy of the
photon is not sufficient to put the electron into a
particular excited energy level, the photon will not
interact with the atom at all!
Yes, a photon of any energy greater than 13.6 eV
will ionize the atom. Any “extra” energy will go
into kinetic energy of the newly liberated electron.
*Q42.3 Answer (a). The 10.5-eV bombarding energy does not match the 10.2-eV excitation energy
required to lift the atom from state 1 to state 2. But the atom can be excited into state 2 and the
bombarding particle can carry off the excess energy.
*Q42.4
(i) b (ii) g From Equations 42.7, 42.8 and 42.9, we have − E = −
Then K = E and U e = −2 E .
489
ke e 2
k e2 k e2
= + e − e = K + Ue .
2r
2r
r
490
*Q42.5
Chapter 42
⎛ 1
1⎞
In ∆E = (13.6 eV ) ⎜ 2 − 2 ⎟ for ∆E > 0 we have absorption and for ∆E < 0 we have emission.
⎝ ni n f ⎠
(a) for ni = 2 and n f = 5, ∆E = 2.86 eV (absorption)
(b)
for ni = 5 and n f = 3,
∆E = −0.967 eV (emission)
(c)
for ni = 7 and n f = 4,
∆E = −0.572 eV (emission)
(d)
for ni = 4 and n f = 7,
∆E = 0.572 eV (absorption)
(i)
In order of energy change, the ranking is a > d > c > b
hc
(ii) E =
so the ranking in order of decreasing wavelength of the associated photon is c = d > b > a.
λ
*Q42.6 (a) Yes. (b) No. The greatest frequency is that of the Lyman series limit. (c) Yes. We can imagine
arbitrarily low photon energies for transitions between adjacent states with n large.
Q42.7
Bohr modeled the electron as moving in a perfect circle, with zero uncertainty in its radial
coordinate. Then its radial velocity is always zero with zero uncertainty. Bohr’s theory violates
the uncertainty principle by making the uncertainty product ∆r ∆pr be zero, less than the
minimum allowable .
2
*Q42.8 (i) d (ii) c and d (iii) b and c. Having n start from 1 and start from zero, with always less than
n, is a way of reminding ourselves that the minimum kinetic energy of a bound quantum particle
is greater than zero and the minimum angular momentum is precisely zero.
Q42.9
Fundamentally, three quantum numbers describe an orbital wave function because we live in
three-dimensional space. They arise mathematically from boundary conditions on the wave
function, expressed as a product of a function of r, a function of q, and a function of f.
Q42.10 Bohr’s theory pictures the electron as moving in a flat circle like a classical particle described
by ΣF = ma. Schrödinger’s theory pictures the electron as a cloud of probability amplitude in
the three-dimensional space around the hydrogen nucleus, with its motion described by a wave
equation. In the Bohr model, the ground-state angular momentum is 1; in the Schrödinger model
the ground-state angular momentum is zero. Both models predict that the electron’s energy
−13.606 eV
is limited to discrete energy levels, given by
with n = 1, 2, 3.
n2
Q42.11 Practically speaking, no. Ions have a net charge and the magnetic force q v × B would deflect
the beam, making it difficult to separate the atoms with different orientations of magnetic
moments.
(
)
Q42.12 The deflecting force on an atom with a magnetic moment is proportional to the gradient of the
magnetic field. Thus, atoms with oppositely directed magnetic moments would be deflected in
opposite directions in an inhomogeneous magnetic field.
Q42.13 If the exclusion principle were not valid, the elements and their chemical behavior would be
grossly different because every electron would end up in the lowest energy level of the atom. All
matter would be nearly alike in its chemistry and composition, since the shell structures of all
elements would be identical. Most materials would have a much higher density. The spectra of
atoms and molecules would be very simple, and there would be very little color in the world.
Atomic Physics
491
Q42.14 In a neutral helium atom, one electron can be modeled as moving in an electric field created by
the nucleus and the other electron. According to Gauss’s law, if the electron is above the ground
state it moves in the electric field of a net charge of +2e − 1e = +1e. We say the nuclear charge is
screened by the inner electron. The electron in a He + ion moves in the field of the unscreened
nuclear charge of 2 protons. Then the potential energy function for the electron is about double
that of one electron in the neutral atom.
Q42.15 The three elements have similar electronic configurations. Each has filled inner shells plus one
electron in an s orbital. Their single outer electrons largely determine their chemical interactions
with other atoms.
Q42.16 Each of the electrons must have at least one quantum number different
from the quantum numbers of each of the other electrons. They can differ
(in ms) by being spin-up or spin-down. They can also differ (in )
in angular momentum and in the general shape of the wave function.
Those electrons with = 1 can differ (in m ) in orientation of angular
momentum—look at Figure Q42.16.
*Q42.17 The M means that the electron falls down into the M shell. The final
principal quantum number is 3. Ma would refer to 4→3 and Mb refers to
5→3. Answers: (i) e (ii) c
FIG. Q42.16
Q42.18 No. Laser light is collimated. The energy generally travels in the same direction. The intensity of
a laser beam stays remarkably constant, independent of the distance it has traveled.
Q42.19 Stimulated emission coerces atoms to emit photons along a specific axis, rather than in the
random directions of spontaneously emitted photons. The photons that are emitted through
stimulation can be made to accumulate over time. The fraction allowed to escape constitutes
the intense, collimated, and coherent laser beam. If this process relied solely on spontaneous
emission, the emitted photons would not exit the laser tube or crystal in the same direction.
Neither would they be coherent with one another.
Q42.20 (a)
The terms “I define” and “this part of the universe” seem vague, in contrast to the precision
of the rest of the statement. But the statement is true in the sense of being experimentally
verifiable. The way to test the orientation of the magnetic moment of an electron is to apply
a magnetic field to it. When that is done for any electron, it has precisely a 50% chance of
being either spin-up or spin-down. Its spin magnetic moment vector must make one of two
S
12
allowed angles with the applied magnetic field. They are given by cos θ = z =
and
S
3 2
−1 2
cos θ =
. You can calculate as many digits of the two angles allowed by “space
3 2
quantization” as you wish.
(b)
This statement may be true. There is no reason to suppose that an ant can comprehend the
cosmos, and no reason to suppose that a human can comprehend all of it. Our experience
with macroscopic objects does not prepare us to understand quantum particles. On the other
hand, what seems strange to us now may be the common knowledge of tomorrow. Looking
back at the past 150 years of physics, great strides in understanding the Universe—from
the quantum to the galactic scale—have been made. Think of trying to explain the
photoelectric effect using Newtonian mechanics. What seems strange sometimes just has
an underlying structure that has not yet been described fully. On the other hand still, it has
been demonstrated that a “hidden-variable” theory, that would model quantum uncertainty
as caused by some determinate but fluctuating quantity, cannot agree with experiment.
492
Chapter 42
*Q42.21 People commonly say that science is difficult to learn. Many other branches of knowledge are
constructed by humans. Learning them can require getting your mind to follow the path of
someone else’s mind. A well-developed science does not mirror human patterns of thought, but
the way nature works. Thus we agree with the view stated in the problem. A scientific discovery
can be like a garbled communication in the sense that it does not explain itself. It can seem
fragmentary and in need of context. It can seem unfriendly in the sense that it has no regard for
human desires or ease of human comprehension. Evolution has adapted the human mind for
catching rabbits and outwitting bison and leopards, but not necessarily for understanding atoms.
Education in science gives the student opportunities for understanding a wide variety of ideas; for
evaluating new information; and for recognizing that extraordinary evidence must be adduced
for extraordinary claims.
SOLUTIONS TO PROBLEMS
Section 42.1
P42.1
(a)
Atomic Spectra of Gases
Lyman series
⎛
1
1⎞
= R ⎜1 − 2 ⎟
ni ⎠
λ
⎝
ni = 2, 3, 4, …
⎛
1
1
1⎞
=
= (1.097 × 10 7 ) ⎜ 1 − 2 ⎟
−9
λ 94.96 × 10
ni ⎠
⎝
(b)
Paschen series:
⎛1
1
1⎞
= R⎜ 2 − 2 ⎟
ni ⎠
λ
⎝3
ni = 5
ni = 4, 5, 6, …
The shortest wavelength for this series corresponds to ni = ∞ for ionization
⎛1 1 ⎞
1
= 1.097 × 10 7 ⎜ − 2 ⎟
λ
⎝ 9 ni ⎠
For ni = ∞, this gives λ = 820 nm
This is larger than 94.96 nm, so this wave
length cannot be associated with the Paschen seriess .
Balmer series:
⎛ 1
1
1⎞
= R⎜ 2 − 2 ⎟
ni ⎠
λ
⎝2
ni = 3, 4, 5, …
⎛1 1 ⎞
1
= 1.097 × 10 7 ⎜ − 2 ⎟
λ
⎝ 4 ni ⎠
with ni = ∞ for ionization, λmin = 365 nm
Once again the shorter given wavelength
cannot be associated with the Balmer series .
Atomic Physics
*P42.2
(a)
493
The fifth excited state must lie above the second excited state by the photon energy
E52 = hf =
hc 6.63 × 10 −34 J ⋅ s 3 × 108 m s
=
= 3.82 × 10 −19 J
λ
520 × 10 −9 m
The sixth excited state exceeds the second in energy by
E62 =
6.63 × 10 −34 J ⋅ s 3 × 108 m s
= 4.85 × 10 −19 J
410 × 10 −9 m
Then the sixth excited state is above the fifth by ( 4.85 − 3.82 ) 10 −19 J = 1.03 × 10 −19 J. In the
6 to 5 transition the atom emits a photon with the infrared wavelength
hc 6.63 × 10 −34 J ⋅ s 3 × 108 m s
=
= 1.94 × 10 −6 m
E62
1.03 × 10 −19 J
λ=
(b)
The same steps solve the symbolic problem.
ECA = ECB + EBA
hc
hc
hc
=
+
λCA λCB λ BA
1
1
1
=
−
λCB λCA λ BA
⎛ 1
1 ⎞
−
λCB = ⎜
⎝ λCA λ BA ⎟⎠
Section 42.2
P42.3
(a)
Early Models of the Atom
For a classical atom, the centripetal acceleration is
a=
1
v2
e2
=
4π ∈0 r 2 me
r
E=−
e2
m v2
e2
+ e =−
4π ∈0 r
2
8π ∈0 r
dE
e2
dr
−1 e2 a 2
=
=
dt 8π ∈0 r 2 dt 6π ∈0 c3
so
⎞
− e2 ⎛
e2
=
3 ⎜
6π ∈0 c ⎝ 4π ∈0 r 2 me ⎟⎠
Therefore,
−
∫
2.00 ×10
T
12π ∈ r m c dr = e
2
−10
m
2
FIG. P42.3
dr
e4
=−
.
dt
12π 2 ∈20 r 2 me2 c3
0
(b)
−1
2
0
2
2 3
e
4
∫ dt
0
12π 2 ∈20 me2 c3 r 3
e4
3
2.00 ×10 −10
= T = 8.46 × 10 −10 s
0
Since atoms last a lot longer than 0.8 ns, the classical laws (fortunately!) do not hold for
systems of atomic size.
494
P42.4
Chapter 42
(a)
The point of closest approach is found when
E = K +U = 0 +
rmin =
(b)
ke ( 2e ) ( 79e )
E
rmin =
or
ke qα qAu
r
(8.99 × 10
9
N ⋅ m 2 C2 ) (158 ) (1.60 × 10 −19 C )
2
( 4.00 MeV ) (1.60 × 10 −13 J MeV )
= 5.68 × 10 −14 m
The maximum force exerted on the alpha particle is
−19
2
2
9
ke qα qAu (8.99 × 10 N ⋅ m C ) (158 ) (1.600 × 10 C )
=
= 11.3 N away from the
2
rmin
(5.68 × 10 −14 m )2
nucleus.
2
Fmax =
Section 42.3
*P42.5
(a)
Bohr’s Model of the Hydrogen Atom
Longest wavelength implies lowest frequency and smallest energy:
the atom falls from
n=3
losing energy
−
The photon frequency is
f =
to
n=2
13.6 eV 13.6 eV
+
= 1.89 eV
32
22
∆E
h
and its wavelength is
λ=
−34
8
c h c ( 6.626 × 10 J ⋅ s ) ( 2.998 ×10 m s ) ⎛
eV
⎞
=
=
⎝ 1.602 × 10 −19 J ⎠
f ∆E
(1.89 eV )
FIG. P42.5
λ = 656 nm
This is the red Balmer-alpha line, which gives its characteristic color to the chromosphere
of the Sun and to photographs of the Orion nebula.
(b)
The biggest energy loss is for an atom to fall from an ionized configuration,
n=∞
to the
n = 2 state
It loses energy
to emit light of wavelength
13.6 eV 13.6 eV
+
= 3.40 eV
22
∞
−34
8
h c ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
λ=
=
= 365 nm
∆E
V ) (1.60 × 10 −19 J eV )
( 3.40 eV
−
This is the Balmer series limit, in the near ultraviolet.
Atomic Physics
P42.6
(a)
ke e 2
me r1
v1 =
r1 = (1) a0 = 0.005 29 nm = 5.29 × 10 −11 m
2
where
(8.99 × 10 N ⋅ m C ) (1.60 × 10 C) = 2.19 × 10 m s
(9.11 × 10 kg) (5.29 × 10 m )
1
1
= m v = ( 9.11 × 10 kg ) ( 2.19 × 10 m s ) = 2.18 × 10 J = 13.6 eV
2
2
(8.99 × 10 N ⋅ m C ) (1.60 × 10 C) = −4.35 × 10 J = −27.2 eV
ke
=−
=−
2
9
v1 =
(b)
K1
2
6
−11
−31
2
e 1
2
6
2
9
−19
2
−18
2
−18
e
(c)
U1
(a)
r22 = ( 0.052 9 nm ) ( 2 ) = 0.212 nm
(b)
−19
2
−31
2
P42.7
495
5.29 × 10 −11 m
r1
2
me v 2 =
me k e e 2
=
r2
(9.11 × 10
−31
kg ) (8.99 × 10 9 N ⋅ m 2 C2 ) (1.60 × 10 −19 C )
2
0.212 × 10 −9 m
me v2 = 9.95 × 10 −25 kg ⋅ m s
(c)
L2 = me v2 r2 = ( 9.95 × 10 −25 kg ⋅ m s ) ( 0.212 × 10 −9 m ) = 2.11 × 10 −34 kg ⋅ m 2 s
(d)
K2 =
(e)
U2 = −
(f)
E2 = K 2 + U 2 = 3.40 eV − 6.80 eV = −3.40 eV
(a)
The batch of excited atoms must make these six transitions to get back to state one: 2 → 1, and
also 3 → 2 and 3 → 1, and also 4 → 3 and 4 → 2 and 4 → 1. Thus, the incoming light
must have just enough energy to produce the 1 → 4 transition. It must be the third line of the
Lyman series in the absorption spectrum of hydrogen. The absorbing atom changes from energy
( m v ) (9.95 × 10 −25 kg ⋅ m s ) = 5.43 × 10 −19 J = 3.40 eV
1
me v22 = e 2 =
2
2 me
2 ( 9.11 × 10 −31 kg )
2
2
(8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10 −19 C) = −1.09 × 10 −18 J = −6.80 eV
ke e 2
=−
0.212 × 10 −9 m
r2
2
*P42.8
13.6 eV
13.6 eV
= −13.6 eV to E f = −
= −0.850 eV
2
1
42
so the incoming photons have wavelength l = c/f = hc/Ephoton
Ei = −
λ=
(6.626 × 10 −34 J ⋅ s ) (3.00 × 108 m s ) ⎛ 1.00 eV ⎞ .
hc
=
⎝ 1.60 × 10 −19 J ⎠
E f − Ei
−0.850 eV − ( −13.6 eV )
= 9.74 × 10 −8 m = 97.4 nm
(b)
The longest of the six wavelengths corresponds to the lowest photon energy, emitted in the
13.6 eV
13.6 eV
4 → 3 transition. Here Ei = −
= −0.850 eV and E f = −
= −1.51 eV,
42
32
so λ =
1240 eV ⋅ nm
hc
=
= 1.888 µ m . This infrared wavelength is part of
E f − Ei −0.850 eV − ( −1.51 eV )
the Paschen series, since the lower state has n = 3.
(c)
The shortest wavelength emitted is the same as the wavelength absorbed: 97.4 nm,
ultraviolet, Lyman series.
496
P42.9
Chapter 42
(a)
The energy levels of a hydrogen-like ion whose charge
number is Z are given by
En = ( −13.6 eV )
Z2
n2
Thus for Helium ( Z = 2 ) , the energy levels are
En = −
(b)
54.4 eV
n2
n = 1, 2, 3, …
For He + , Z = 2, so we see that the ionization energy (the
energy required to take the electron from the n = 1 to
the n = ∞ state) is
FIG. P42.9
( −13.6 eV ) ( 2 )
= 54.4 eV
(1)2
2
E = E∞ − E1 = 0 −
*P42.10 (a)
The photon has energy 2.28 eV.
13.6 eV
= 3.40 eV is required to ionize a hydrogen atom from state n = 2. So while
22
the photon cannot ionize a hydrogen atom pre-excited to n = 2, it can ionize a hydrogen
13.6 eV
atom in the n = 3 state, with energy −
= −1.51 eV.
32
And
(b)
The electron thus freed can have kinetic energy K e = 2.28 eV − 1.51 eV = 0.769 eV =
2 ( 0.769 ) (1.60 × 10 −19 ) J
Therefore v =
P42.11
9.11 × 10 −31 kg
1
me v 2 .
2
ms .
= 520 km
Let r represent the distance between the electron and the positron. The two move in a circle of
r
radius
around their center of mass with opposite velocities. The total angular momentum of
2
the electron-positron system is quantized according to
Ln =
where
m vr m vr
+
= n
2
2
n = 1, 2, 3, …
For each particle, ΣF = ma expands to
We can eliminate v =
n
to find
mr
So the separation distances are
The orbital radii are
ke e 2
,
2r
ke e2 2 mn 2 2
=
r
m2r 2
r=
2n 2 2
= 2 a0 n 2 =
mke e2
(1.06 × 10
−10
m ) n2
r
= a0 n 2 , the same as for the electron in hydrogen.
2
The energy can be calculated from
Since m v 2 =
ke e 2 m v 2
=
r2
r2
E = K +U =
E=
1
1
k e2
mv 2 + mv 2 − e
2
2
r
6.80 eV
ke e 2 ke e 2
k e2 − k e2
−
=− e = e 2 = −
2r
2r
4 a0 n
r
n2
Atomic Physics
*P42.12 (a)
From the Bohr theory we have for the speed of the electron v =
The period of its orbital motion is T =
r=
497
n
.
me r
2π r 2π rme r
=
. Substituting the orbital radius
n
v
2π me n 4 4
2π 3 3
n22
gives
T
=
=
n . Thus we have the periods determined in
nme2 ke2 e4 me ke2 e4
me k e e 2
terms of the ground-state period
t0 =
(b)
In the n = 2 state the period is (153 × 10–18 s)23 = 1.22 × 10–15 s so the number of orbits
completed in the excited state is 10 × 10–6 s/1.22 × 10–15 s = 8.18 × 109 revolutions .
(c)
Its lifetime in electron years is comparable to the lifetime of the Sun in Earth years, so we
can think of it as a long time.
Section 42.4
P42.13
2π 3
2π (1.055 × 10 −34 J ⋅ s)3
=
= 153 as
me ke2 e4 9.11 × 10 −31 kg(8.99 × 10 9 N ⋅ m 2 /C2 )2 (1.6 × 10 −19 C)4
The Quantum Model of the Hydrogen Atom
The reduced mass of positronium is less than hydrogen, so the photon energy will be less for
positronium than for hydrogen. This means that the wavelength of the emitted photon will be
longer than 656.3 nm. On the other hand, helium has about the same reduced mass but more
charge than hydrogen, so its transition energy will be larger, corresponding to a wavelength
shorter than 656.3 nm.
All the factors in the given equation are constant for this problem except for the reduced mass
and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the
transition can be found simply from the ratio of mass and charge variables.
m p me
For hydrogen,
µ=
Its wavelength is
λ = 656.3 nm
(a)
m p + me
µ=
For positronium,
≈ me
The photon energy is
∆E = E3 − E2
where
λ=
me me
m
= e
2
me + me
c hc
=
f ∆E
so the energy of each level is one half as large as in hydrogen, which we could call
“protonium.” The photon energy is inversely proportional to its wavelength, so for
positronium,
λ32 = 2 ( 656.3 nm ) = 1.31 µ m (in the infrared region)
(b)
For He +,
µ ≈ me , q1 = e , and q2 = 2e
so the transition energy is 2 = 4 times larger than hydrogen
2
Then,
656 ⎞
nm = 164 nm (in the ultraviolet region)
λ32 = ⎛
⎝ 4 ⎠
498
Chapter 42
*P42.14 (a)
For a particular transition from ni to n f , ∆EH = −
∆ED = −
µ D ke 2 e 4
2 2
By division,
(b)
µ H ke 2 e 4
2 2
⎛ 1
1 ⎞ hc
⎜ n 2 − n 2 ⎟ = λ and
⎝ f
i ⎠
H
⎛ 1
me m p
me mD
1 ⎞ hc
⎜ n 2 − n 2 ⎟ = λ where µH = m + m and µD = m + m .
⎝ f
e
p
i ⎠
D
e
D
⎛µ ⎞
⎛
∆EH µH λ D
µ ⎞
=
=
or λ D = ⎜ H ⎟ λ H . Then, λH − λD = ⎜ 1 − H ⎟ λH .
∆ED µD λ H
µD ⎠
⎝ µD ⎠
⎝
µH ⎛ me m p ⎞ ⎛ me + mD ⎞ (1.007 276 u ) ( 0.000 549 u + 2.013 553 u )
=
=
= 0.999 728
µD ⎜⎝ me + m p ⎟⎠ ⎜⎝ me mD ⎟⎠ ( 0.000 549 u + 1.007 276 u ) ( 2.013 553 u )
λH − λD = (1 − 0.999 728 ) ( 656.3 nm ) = 0.179 nm
*P42.15 (a)
∆x ∆p ≥
so if ∆x = r, ∆p ≥
2
2r
( ∆p )
p2
2
, we find K =
≈
=
r
2 me
2 me
2 me r 2
2
(b)
Arbitrarily choosing ∆p ≈
U=
(c)
2
ke e 2
− ke e 2
, so E = K + U ≈
−
r
r
2 me r 2
To minimize E,
2
2
dE
ke e 2
=−
+
=
0
→
=
= a0
r
dr
me r 3
r2
me k e e 2
(the Bohr radius)
2
Then, E =
⎛ m k e2 ⎞
m k 2 e4
2 ⎛ me k e e 2 ⎞
− ke e2 ⎜ e 2e ⎟ = − e e2 = −13.6 eV . With our particular
2
⎜
⎟
⎝ ⎠
2 me ⎝ ⎠
2
choice for the momentum uncertainty as double its minimum possible value, we find
precisely the Bohr results for the orbital radius and for the ground-state energy.
Section 42.5
P42.16
The Wave Functions of Hydrogen
ψ 1s ( r ) =
1
e− r a0 is the ground state
π a03
hydrogen wave function.
P1s ( r ) =
4 r 2 −2 r a0
e
is the ground state radial
a03
probability distribution function.
FIG. P42.16
Atomic Physics
P42.17
(a)
∫ψ
2
∞
⎛ 1 ⎞ ∞ 2 −2 r a0
2
dV = 4π ∫ ψ r 2 dr = 4π ⎜
r e
dr
⎝ π a03 ⎟⎠ ∫0
0
∞
a02 ⎞ ⎤ ⎛ 2 ⎞ ⎛ a02 ⎞
2 ⎡ −2 r a0 ⎛ 2
ψ
dV
=
−
e
r
+
a
r
+
Using integral tables,
= 1
−
⎥ = −
⎢
0
⎜⎝
∫
a02 ⎣
2 ⎟⎠ ⎦ 0 ⎜⎝ a02 ⎟⎠ ⎜⎝ 2 ⎟⎠
so the wave function as given is normalized.
2
3 a0 2
(b)
∫
Pa0 / 2→3a0 / 2 = 4π
a0 2
⎛ 1 ⎞
2
ψ r 2 dr = 4π ⎜ 3 ⎟
⎝ π a0 ⎠
3 a0 2
∫
r 2 e−2 r a0 dr
a0 2
Again, using integral tables,
3 a0 2
Pa0 /2→3a0 / 2
P42.18
⎛
a2 ⎞ ⎤
2 ⎡
= − 2 ⎢ e−2 r a0 ⎜ r 2 + a0 r + 0 ⎟ ⎥
⎝
a0 ⎣
2 ⎠ ⎦a
0
=−
2
ψ=
1
1
r − r 2 a0
e
32
3 ( 2a0 ) a0
so
Pr = 4π r 2 ψ 2 = 4π r 2
Set
⎤
⎛ 1⎞
dP
4π ⎡ 3 − r a0
=
+ r 4 ⎜ − ⎟ e− r a0 ⎥ = 0
⎢ 4r e
dr 24 a05 ⎣
a
⎝ 0⎠
⎦
2
2 ⎡ −3 ⎛ 17a02 ⎞
−1 ⎛ 5a0 ⎞ ⎤
−
e
e
⎥ = 0.497
⎢
⎝⎜ 4 ⎟⎠ ⎦
a02 ⎣ ⎜⎝ 4 ⎟⎠
r 2 − r a0
e
24 a05
Solving for r, this is a maximum at r = 4 a0 .
P42.19
ψ=
1
πa
3
0
2 dψ
2
−2
e− r a0 = −
ψ
=
ra0
r dr r π a05
e− r a0
d 2ψ
1
1
e− r a0 = 2 ψ
=
2
7
a0
dr
π a0
Substitution into the Schrödinger equation to test the validity of the solution yields
−
2 ⎛ 1
2 ⎞
e2
−
ψ−
ψ = Eψ
2
⎜
⎟
2 me ⎝ a0 ra0 ⎠
4π ∈0 r
But
a0 =
so
−
2 ( 4π ∈0 )
me e 2
e2
=E
8π ∈0 a0
E=−
or
ke e 2
2a0
This is true, so the Schrödinger equation is satisfied.
P42.20
The hydrogen ground-state radial probability density is
P ( r ) = 4π r 2 ψ 1s =
2
⎛ 2r ⎞
4r 2
exp ⎜ − ⎟
3
a0
⎝ a0 ⎠
The number of observations at 2a0 is, by proportion
P ( 2 a0 )
( 2a0 ) e−4 a0 a0 = 1 000 (16) e−3 = 797 times
N = 1 000
= 1 000
P ( a0 2 )
( a0 2)2 e− a0 a0
2
499
500
Chapter 42
Section 42.6
Physical Interpretation of the Quantum Numbers
Note: Problems 31 and 36 in Chapter 29 and Problem 62 in Chapter 30 can be assigned with this section.
P42.21
(a)
In the 3d subshell, n = 3 and = 2,
3
we have n
2
m
+2
ms +1/2
3
2
+2
–1/2
3
2
+1
+1/2
3
2
+1
–1/2
3
2
0
+1/2
3
2
0
–1/2
3
1
0
–1/2
3
1
–1
+1/2
3
1
–1
–1/2
3
2
–1
+1/2
3
2
–1
–1/2
3
2
–2
+1/2
3
2
–2
–1/2
(A total of 10 states)
(b)
In the 3p subshell, n = 3 and = 1,
3
we have n
1
m
+1
ms +1/2
3
1
+1
–1/2
3
1
0
+1/2
(A total of 6 states)
P42.22
−34
6 = 2.58 × 10 J ⋅ s
(a)
For the d state, = 2,
L=
(b)
For the f state, = 3,
L = ( + 1) =
*P42.23 (a)
(b)
−34
12 = 3.65 × 10 J ⋅ s
The problem: Find the orbital quantum number of a hydrogen atom in a state in which it
has orbital angular momentum 4.714 × 10−34 J ⋅ s.
⎛ 6.626 × 10 −34 ⎞
4.714 × 10 −34 = ( + 1) ⎜
⎟⎠
⎝
2π
The solution: L = ( + 1)
( + 1) =
( 4.714 × 10 ) ( 2π )
(6.626 × 10 )
−34 2
−34 2
2
= 1.998 × 101 ≈ 20 = 4 ( 4 + 1)
so the orbital quantum number is = 4 .
P42.24
The 5th excited state has n = 6, energy
−13.6 eV
= −0.378 eV.
36
The atom loses this much energy:
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
= 1.14 eV
λ (1 090 × 10 −9 m ) (1.60 × 10 −19 J eV )
to end up with energy
−0.378 eV − 1.14 eV = −1.52 eV
which is the energy in state 3:
−
13.6 eV
= −1.51 eV
33
While n = 3, can be as large as 2, giving angular momentum
( + 1) =
6 .
Atomic Physics
P42.25
(a)
n = 1:
For n = 1, = 0, m = 0, ms = ±
n
m
1
1
0
0
0
0
501
1
2
ms
–1/2
+1/2
Yields 2 sets; 2n 2 = 2 (1) = 2
2
(b)
n = 2:
For n = 2,
we have
n
m
ms
2
2
2
2
0
1
1
1
0
–1
0
1
±1/2
±1/2
±1/2
±1/2
Yields 8 sets; 2n 2 = 2 ( 2 ) = 8
2
Note that the number is twice the number of m values. Also, for each there are ( 2 + 1)
different m values. Finally, can take on values ranging from 0 to n − 1.
n −1
number = ∑ 2 ( 2 + 1)
So the general expression is
0
2 + 6 + 10 + 14 …
The series is an arithmetic progression:
n
[ 2a + ( n − 1) d ]
2
n
number = [ 4 + ( n − 1) 4 ] = 2n 2
2
number =
the sum of which is
where a = 2, d = 4:
P42.26
(c)
n = 3:
2 (1) + 2 ( 3) + 2 ( 5) = 2 + 6 + 10 = 18
2n 2 = 2 ( 3) = 18
(d)
n = 4:
2 (1) + 2 ( 3) + 2 ( 5) + 2 ( 7 ) = 32
2n 2 = 2 ( 4 ) = 32
(e)
n = 5:
32 + 2 ( 9 ) = 32 + 18 = 50
2n 2 = 2 ( 5) = 50
n = 3 and = 2
Therefore,
L = ( + 1) =
m can have the values
–2, –1, 0, 1, and 2
Using the relation
we find the possible values of q
2
2
For a 3d state,
so
2
−34
6 = 2.58 × 10 J ⋅ s
Lz can have the values − 2, − , 0, and 2 .
cos θ =
Lz
L
145°, 114°, 90.0°, 65.9°, and 35.3°
502
Chapter 42
*P42.27 (a)
ρ=
Density of a proton:
m
1.67 × 10 −27 kg
17
=
kg m 3
3 = 3.99 × 10
V ( 4 3) π (1.00 × 10 −15 m )
(b)
⎛ 3m ⎞
Size of model electron: r = ⎜
⎝ 4π ρ ⎟⎠
(c)
Moment of inertia:
⎛ 3 ( 9.11 × 10 −31 kg ) ⎞
=⎜
17
3 ⎟
⎝ 4π ( 3.99 × 10 kg m ) ⎠
13
= 8.17 × 10 −17 m
2
2 2 2
mr = ( 9.11 × 10 −31 kg ) (8.17 × 10 −17 m )
5
5
= 2.43 × 10 −63 kg ⋅ m 2
I=
Lz = I ω =
v=
Therefore,
(d)
13
Iv
=
2
r
−34
−17
r ( 6.626 × 10 J ⋅ s ) (8.17 × 10 m )
=
= 1.77 × 1012 m s
2I
2π ( 2 × 2.43 × 10 −63 kg ⋅ m 2 )
This is 5.91 × 10 3 times larger than the speed of light. So the spinning-solid-ball model
of an electron with spin angular momentum is absurd.
P42.28
In the N shell, n = 4. For n = 4, can take on values of 0, 1, 2, and 3. For each value of , m
can be − to in integral steps. Thus, the maximum value for m is 3. Since Lz = m , the
maximum value for Lz is Lz = 3 .
P42.29
The 3d subshell has = 2, and n = 3. Also, we have s = 1.
Therefore, we can have n = 3, = 2; m = −2, − 1, 0, 1, 2; s = 1; and ms = −1, 0, 1
leading to the following table:
n
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
m
–2
–2
–2
–1
–1
–1
0
0
0
1
1
1
2
2
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
–1
0
1
–1
0
1
–1
0
1
–1
0
1
–1
0
1
s
ms
Atomic Physics
Section 42.7
The Exclusion Principle and the Periodic Table
P42.30
1s 2 2 s 2 2 p 4
(a)
(b)
For the 1s electrons,
n = 1, = 0, m = 0,
For the two 2s electrons,
n = 2, = 0, m = 0,
1
2
1
ms = +
2
1
ms = +
2
ms = +
For the four 2p electrons, n = 2; = 1; m = −1, 0, or 1; and
and
and
or
503
1
2
1
−
2
1
−
2
−
So one possible set of quantum numbers is
n
m
1
0
0
1
0
0
ms
1
2
– 12
2
0
0
1
2
2
0
0
– 12
2
1
–1
2
1
0
2
1
1
2
1
–1
1
2
1
2
1
2
– 12
P42.31
The 4s subshell fills first , for potassium and calcium, before the 3d subshell starts to fill for
scandium through zinc. Thus, we would first suppose that [ Ar ] 3d 4 4 s 2 would have lower energy
than [ Ar ] 3d 5 4 s1. But the latter has more unpaired spins, six instead of four, and Hund’s rule
suggests that this could give the latter configuration lower energy. In fact it must, for [ Ar ] 3d 5 4 s1
is the ground state for chromium.
P42.32
Electronic configuration:
2
2
6
⎣⎡1s 2 s 2 p ⎤⎦
+3s1
→
Na 11
+3s 2
→
Mg12
+3s 2 3 p1
→
Al13
+3s 2 3 p 2
→
Si14
+3s 2 3 p 3
→
P15
+3s 2 3 p 4
→
S16
+3s 2 3 p 5
→
Cl17
+3s 2 3 p 6
→
Ar18
→
K19
2
2
6
2
6
1
⎣⎡1s 2 s 2 p 3s 3 p ⎤⎦ 4 s
P42.33
Sodium to Argon
In the table of electronic configurations in the text, or on a periodic table, we look for the element
whose last electron is in a 3p state and which has three electrons outside a closed shell. Its
electron configuration then ends in 3s 2 3 p1. The element is aluminum .
504
P42.34
Chapter 42
(a)
For electron one and also for electron two, n = 3 and = 1 . The possible states are listed
here in columns giving the other quantum numbers:
electron
m
1
1
1
1
1
one
ms
1
2
1
2
1
2
1
2
1
2
electron
m
1
0
0
–1
–1
1
0
two
ms −
1
2
−
1
2
1
2
1
2
−
electron
m
–1
one
ms −
electron
m
1
two
ms
1
2
1
2
0
1
2
0
−
1
2
0
0
0
1
1
1
1
1
− − − −
2
2
2
2
2
1
−
1
2
1
−
1
1
1
1
1
1
1
1
1
− − − −
2
2
2
2
2
0
0
0
0
1
2
1
2
1
2
1
2
1
2
1
0
–1
–1
1
2
−
–1
–1
–1
–1
1
1
2
1
2
−
1
2
1
2
−
–1
–1
–1
–1
–1
–1
1
2
1
2
1
2
1
2
1
2
−
1
0
0
–1
1
1
1
−
2
2
1
2
0
–1
–1
1
1
2
1
2
−
1
2
1
2
−
1
2
0
0
1
2
−
1
1
−
2
2
–1
1
2
1
1
1
1
1
− − − −
2
2
2
2
2
1
−
1
2
0
1
2
0
−
1
2
–1
1
2
There are thirty allowed states, since electron one can have any of three possible values for
m for both spin up and spin down, amounting to six states, and the second electron can
have any of the other five states.
(b)
P42.35 (a)
(b)
Were it not for the exclusion principle, there would be 36 possible states, six for each
electron independently.
n+
1
2
3
4
subshell
1s
2s
2p, 3s
3p, 4s
5
6
7
3d, 4p, 5s 4d , 5p, 6s 4f , 5d, 6p, 7s
Z = 15:
Filled subshells:
Valence subshell:
Prediction:
Element is phosphorus,
1s, 2 s, 2 p, 3s (12 electrons)
3 electrons in 3p subshell
Valence = −3 or +5
Valence = −3 or +5 (Prediction correct)
Z = 47:
Filled subshells:
1s, 2 s, 2 p, 3s, 3 p, 4 s, 3d , 4 p, 5s
(38 electrons)
9 electrons in 4d subshell
Valence = −1
(Prediction fails)
Valence is +1
Outer subshell:
Prediction:
Element is silver,
Z = 86:
Filled subshells:
Prediction
Element is radon, inert
1s, 2 s, 2 p, 3s, 3 p, 4 s, 3d , 4 p, 5s, 4 d , 5 p, 6 s,
4 f , 5d , 6 p (86 electrons)
Outer subshell is full: inert gas
(Prediction correct)
Atomic Physics
P42.36
505
Listing subshells in the order of filling, we have for element 110,
1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 6 5s 2 4 d 10 5 p 6 6 s 2 4 f 14 5d 10 6 p 6 7 s 2 5 f 14 6d 8
In order of increasing principal quantum number, this is
1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4 d 10 4 f 14 5s 2 5 p 6 5d 10 5 f 14 6 s 2 6 p 6 6d 8 7 s 2
P42.37
In the ground state of sodium, the outermost electron is in an s state. This state is spherically
symmetric, so it generates no magnetic field by orbital motion, and has the same energy no matter
whether the electron is spin-up or spin-down. The energies of the states 3p ↑ and 3p ↓ above 3s
hc
hc
are hf1 =
and hf2 =
.
λ
λ2
The energy difference is
⎛ 1
1⎞
2 µ B B = hc ⎜ − ⎟
⎝ λ1 λ2 ⎠
so
B=
1⎞
hc ⎛ 1
−
2 µ B ⎜⎝ λ1 λ2 ⎟⎠
(6.63 × 10 J ⋅ s) (3 × 10 m s) ⎛
1
⎜⎝
588.995 × 10
2 ( 9.27 × 10
J T)
−34
=
8
−24
−9
m
−
1
⎞
⎟
589.592 × 10 −9 m ⎠
B = 18.4 T
Section 42.8
More on Atomic Spectra: Visible and X-ray
P42.38
n = 3, = 0, m = 0
(a)
n = 3, = 1, m = −1, 0, 1
For n = 3, = 2, m = −2, − 1, 0, 1, 2
(b)
ψ 300 corresponds to E300 = −
Z 2 E0
22 (13.6 )
= −6.05 eV .
=−
2
32
n
ψ 31−1 , ψ 310 , ψ 311 have the same energy since n is the same.
ψ 32− 2 , ψ 32−1 , ψ 320 , ψ 321 , ψ 322 have the same energy since n is the same.
All states are degenerate.
506
Chapter 42
*P42.39 For the 3p state, En =
−13.6 eV Z eff2
For the 3d state −1.5 eV =
n2
becomes −3.0 eV =
−13.6 eV Z eff2
so
32
−13.6 eV Z eff2
32
so
Z efff = 1.4
Z efff = 1.0
When the outermost electron in sodium is promoted from the 3s state into a 3p state, its wave
function still overlaps somewhat with the ten electrons below it. It therefore sees the +11e nuclear
charge not fully screened, and on the average moves in an electric field like that created by a
particle with charge +11e – 9.6e = 1.4e. When this valence electron is lifted farther to a 3p state,
it is essentially entirely outside the cloud of ten electrons below it, and moves in the field of a net
charge +11e – 10e = 1e.
*P42.40 (a)
Picture all of the energy of an electron after its acceleration going into producing a single
hc
photon. Then we have E =
= e∆V and
λ
∆V =
hc 6.626 × 10 −34 J ⋅ s 2.998 × 108 m/s 1240 V ⋅ nm
=
=
λ
λ
1.602 × 10 −19 J/eV
λ
(b)
The potential difference is inversely proportional to the wavelength.
(c)
Yes. It predicts a minimum wavelength of 33.5 pm when the accelerating voltage is 37 keV,
in agreement with the minimum wavelength in the figure.
(d)
Yes, but it might be unlikely for a very high-energy electron to stop in a single interaction
to produce a high-energy gamma ray; and it might be difficult to observe the very lowintensity radio waves produced as bremsstrahlung by low-energy electrons. The potential
difference goes to infinity as the wavelength goes to zero. The potential difference goes to
zero as the wavelength goes to infinity.
P42.41 Following Example 42.5
Eγ =
3
( 42 − 1)2 (13.6 eV ) = 1.71 × 10 4 eV = 2.74 × 10 −15 J
4
f = 4.14 × 1018 Hz
λ = 0.072 5 nm
and
P42.42
E=
For
hc 1 240 eV ⋅ nm 1.240 keV ⋅ nm
=
=
λ
λ
λ
λ1 = 0.018 5 nm,
E = 67.11 keV
λ2 = 0.020 9 nm,
E = 59.4 keV
λ3 = 0.021 5 nm,
E = 57.7 keV
FIG. P42.42
The ionization energy for the K shell is 69.5 keV, so the ionization energies for the other shells are:
L shell = 11.8 keV
M shell = 10.1 keV
N shell = 2.39 keV
Atomic Physics
P42.43
The K β x-rays are emitted when there is a vacancy in the ( n = 1) K shell and an electron from the
(n = 3) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and
by one in its final state.
hc
13.6 ( Z − 9 )
13.6 ( Z − 1)
eV +
eV
=−
λ
32
12
2
(6.626 × 10
( 0.152 × 10
2
J ⋅ s ) ( 3.00 × 108 m s )
−34
−9
m ) (1.60 × 10
−19
⎛ Z 2 18 Z 81
⎞
= (13.6 eV ) ⎜ −
+
− + Z 2 − 2 Z + 1⎟
⎝ 9
⎠
9
9
J eV )
⎛ 8Z 2
⎞
8.17 × 10 3 eV = (13.6 eV ) ⎜
− 8⎟
⎝ 9
⎠
8Z 2
−8
9
so
601 =
and
Z = 26
Iron
Section 42.9
Spontaneous and Stimulated Transitions
Section 42.10
Lasers
P42.44
E4 − E3 = ( 20.66 − 18.70 ) eV = 1.96 eV =
The photon energy is
λ=
P42.45
(6.626 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s )
1.96 (1.60 × 10 −19 J )
f =
⎛ 1.60 × 10 −19 C ⎞ ⎛ 1 J ⎞
E
0.117 eV
13 −1
=
−34
⎟⎠ ⎝ 1 V ⋅ C ⎠ = 2.82 × 10 s
h 6.630 × 10 J ⋅ s ⎜⎝
e
λ=
c 3.00 × 108 m s
=
= 10.6 µ m , infrared
f
2.82 × 1013 s −1
(3.00 × 10 J )
(1.00 × 10 s ) ⎡⎣π (15.0 × 10
−3
P42.46
P42.47
507
(a)
I=
(b)
(3.00 × 10 J )
−9
−3
( 0.600 × 10
(30.0 × 10
−9
−6
−6
m)
m)
m) ⎤
⎦
2
= 4.24 × 1015 W m 2
2
2
= 1.20 × 10 −12 J = 7.50 MeV
E = P ∆t = (1.00 × 10 6 W ) (1.00 × 10 −8 s ) = 0.010 0 J
Eγ = hf =
N=
−34
8
hc ( 6.626 × 10 ) ( 3.00 × 10 )
=
J = 2.86 × 10 −19 J
λ
694.3 × 10 −9
0.010 0
E
=
= 3.49 × 1016 photons
Eγ 2.86 × 10 −19
hc
λ
= 633 nm
508
P42.48
Chapter 42
(a)
−E
N3 N g e 3
=
N 2 N g e − E2
( kB ⋅300 K )
( kB ⋅300 K )
= e−( E3 − E2 ) ( kB ⋅300 K ) = e− hc λ ( kB ⋅300 K )
where l is the wavelength of light radiated in the 3 → 2 transition.
N3
−( 6.63×10 −34 J⋅s )( 3×108 m s ) ( 632.8 ×10 −9 m )(1.38 ×10 −23 J K )( 300 K )
=e
N2
N3
= e−75.9 = 1.07 × 10 −33
N2
(b)
Nu
= e−( Eu − E ) kBT
N
where the subscript u refers to an upper energy state and the subscript to a lower energy state.
Since Eu − E = Ephoton =
hc
λ
Nu
= e− hc λ kBT
N
Thus, we require
1.02 = e− hc λ kBT
or
ln (1.02 ) = −
T =−
(6.63 × 10
(632.8 × 10
−9
−34
J ⋅ s ) ( 3 × 108 m s )
m ) (1.38 × 10 −23 J K ) T
2.28 × 10 4
= −1.15 × 10 6 K
ln (1.02 )
A negative-temperature state is not achieved by cooling the system below 0 K, but by
heating it above T = ∞, for as T → ∞ the populations of upper and lower states approach
equality.
(c)
Because Eu − E > 0, and in any real equilibrium state T > 0 ,
e−( Eu − E ) kBT < 1
and
Nu < N
Thus, a population inversion cannot happen in thermal equilibrium.
P42.49
(a)
The light in the cavity is incident perpendicularly on the mirrors,
although the diagram shows a large angle of incidence for clarity.
We ignore the variation of the index of refraction with wavelength.
To minimize reflection at a vacuum wavelength of 632.8 nm,
the net phase difference between rays (1) and (2) should be 180°.
There is automatically a 180° shift in one of the two rays upon
reflection, so the extra distance traveled by ray (2) should be
one whole wavelength:
2t =
t=
(b)
1
2
SiO2
FIG. P42.49
λ
n
λ 632.8 nm
=
= 217 nm
2n 2 (1.458 )
The total phase difference should be 360°, including contributions of 180° by reflection and
180° by extra distance traveled
2t =
t=
λ
2n
543 nm
λ
=
= 93.1 nm
4 n 4 (1.458 )
Atomic Physics
509
Additional Problems
P42.50
(a)
Using the same procedure that was used in the Bohr model of the hydrogen atom, we
apply Newton’s second law to the Earth. We simply replace the Coulomb force by the
gravitational force exerted by the Sun on the Earth and find
G
MS M E
v2
= ME
2
r
r
(1)
where v is the orbital speed of the Earth. Next, we apply the postulate that angular
momentum of the Earth is quantized in multiples of :
( n = 1,
M E vr = n
2, 3, …)
Solving for v gives
v=
n
ME r
(2)
Substituting (2) into (1), we find
r=
(b)
n22
GM S M E2
(3)
Solving (3) for n gives
n = GM S r
ME
(4)
Taking M S = 1.99 × 10 30 kg , and M E = 5.98 × 10 24 kg, r = 1.496 × 1011 m,
G = 6.67 × 10 −11 Nm 2 kg2 , and = 1.055 × 10 −34 Js, we find
n = 2.53 × 10 74
(c)
We can use (3) to determine the radii for the orbits corresponding to the quantum numbers
n and n + 1:
rn =
n22
GM S M E2
and
rn+1 =
( n + 1)2 2
GM S M E2
Hence, the separation between these two orbits is
∆r =
2
2
2
2
⎡
⎤
)
=
n
+
1
−
n
(
⎦ GM M 2 ( 2n + 1)
GMS ME2 ⎣
E
S
Since n is very large, we can neglect the number 1 in the parentheses and express the
separation as
∆r ≈
2
( 2n ) = 1.18 × 10 −63 m
GM S M E2
This number is much smaller than the radius of an atomic nucleus ( ~10 −15 m ) , so the
distance between quantized orbits of the Earth is too small to observe.
510
P42.51
Chapter 42
(a)
∆E =
−19
−34
eB 1.60 × 10 C ( 6.63 × 10 J ⋅ s ) ( 5.26 T ) ⎛ N ⋅ s ⎞ ⎛ kg ⋅ m ⎞
= 9.75 × 10 −23 J
=
⎝ T ⋅ C ⋅ m ⎠ ⎝ N ⋅ s2 ⎠
me
2π ( 9.11 × 10 −31 kg )
= 609 µ eV
(b)
kBT = (1.38 × 10 −23 J K ) (80 × 10 −3 K ) = 1.10 × 10 −24 J = 6.9 µ eV
(c)
f=
∆E
9.75 × 10 −23 J
=
= 1.47 × 1011 Hz
6.63 × 10 −34 J ⋅ s
h
λ=
c
3 × 108 m s
=
= 2.04 × 10 −3 m
f 1.47 × 1011 Hz
∞
∞
(a)
= ∫ P1s ( r ′ ) dr ′ =
Probability
r
∞
⎡ ⎛ 2r ′ 2 2r ′ ⎞ −2r ′ a0 ⎤
4
r ′ 2 e−2 r ′ a0 dr ′ = ⎢ − ⎜ 2 +
+ 1⎟ e
⎥
3 ∫
a0
a0 r
⎠
⎣ ⎝ a0
⎦r
using integration by parts, we find
⎛ 2r 2 2r ⎞ −2 r a0
= ⎜ 2 +
+ 1⎟ e
a0
⎝ a0
⎠
(b)
Probability Curve for Hydrogen
1.2
1
Probability
P42.52
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
r /a0
FIG. P42.52
(c)
The probability of finding the electron inside or outside the sphere of radius r is
⎛ 2r 2 2r ⎞
2r
1
∴ ⎜ 2 + + 1⎟ e −2r a0 =
or z 2 + 2 z + 2 = ez where z =
a0
2
⎝ a0 a0 ⎠
1
.
2
One can home in on a solution to this transcendental equation for r on a calculator,
the result being r = 1.34 a0 to three digits.
Atomic Physics
P42.53
hf = ∆ E =
4π 2 me ke2 e 4 ⎛ 1
1⎞
⎜⎝ ( n − 1)2 − n 2 ⎟⎠
2h 2
f=
2π 2 me ke2 e 4 ⎛ 2n − 1 ⎞
⎜⎝ ( n − 1)2 n 2 ⎟⎠
h3
2π 2 me ke2 e 4 2
h3
n3
As n approaches infinity, we have f approaching
f=
The classical frequency is
r=
where
v
1
=
2π r 2π
ke e 2 1
me r 3 2
n2h2
4π me ke e 2
Using this equation to eliminate r from the expression for f, we find f =
in agreement with the Bohr result for large n.
P42.54
(a)
2π 2 me ke2 e 4 2
h3
n3
The energy difference between these two states is equal to the energy that is absorbed.
Thus,
P42.55
511
3
kBT or
2
E = E2 − E1 =
( −13.6 eV ) ( −13.6 eV )
4
−
1
= 10.2 eV = 1.63 × 10 −18 J
2 (1.63 × 10 J )
2E
=
= 7.88 × 10 4 K
3kB 3 (1.38 × 10 −23 J K )
−18
(b)
E=
T=
(a)
The energy of the ground state is:
E1 = −
hc
λseries limit
=−
1 240 eV ⋅ nm
152.0 nm
= −8.16 eV
From the wavelength of the Lyman a line:
E2 − E1 =
hc 1 240 nm ⋅ eV
=
= 6.12 eV
λ
202.6 nm
E2 = E1 + 6.12 eV = −2.04 eV
1 240 nm ⋅ eV
= 7.26 eV
170.9 nm
The wavelength of the Lyman b line gives:
E3 − E1 =
so
E3 = −0.902 eV
Next, using the Lyman g line gives:
E4 − E1 =
and
E4 = −0.508 eV
From the Lyman d line,
E5 − E1 =
so
E5 = −0.325 eV
1 240 nm ⋅ eV
= 7.65 eV
162.1 nm
1 240 nm ⋅ eV
= 7.83 eV
158.3 nm
continued on next page
512
Chapter 42
(b)
For the Balmer series,
hc
1 240 nm ⋅ eV
= Ei − E2 , or λ =
λ
Ei − E2
For the a line, Ei = E3 and so
λa =
1 240 nm ⋅ eV
( −0.902 eV ) − ( −2.04 eV )
= 1 090 nm
Similarly, the wavelengths of the b line, g line, and the short wavelength limit are found to be:
811 nm , 724 nm , and 609 nm .
(c)
Computing 60.0% of the wavelengths of the spectral lines shown on the energy-level
diagram gives:
0.600 ( 202.6 nm ) = 122 nm , 0.600 (170.9 nm ) = 103 nm ,
0.600 (162.1 nm ) = 97.3 nm , 0.600 (158.3 nm ) = 95.0 nm ,
and 0.600 (152.0 nm ) = 91.2 nm
These are seen to be the wavelengths of the a, b, g, and d lines as well as the short
wavelength limit for the Lyman series in Hydrogen.
(d)
The observed wavelengths could be the result of Doppler shift when the source moves
away from the Earth. The required speed of the source is found from
1 − (v c )
f′ λ
= 0.600
=
=
λ′
f
1 + (v c )
*P42.56 (a)
−34
8
hc
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
=
= 4.87 × 10 −7 m = 487 nm
Eγ
∆E
( 2.55 eV ) (1.60 × 10 −19 J eV )
Since momentum must be conserved, the photon and the atom go in opposite directions
h
with equal magnitude momenta. Thus, p = matom v = or
λ
v=
P42.57
v = 0.471c
1
1
The energy emitted by the atom is ∆E = E4 − E2 = −13.6 eV ⎛ 2 − 2 ⎞ = 2.55 eV . The
⎝4
2 ⎠
wavelength of the photon produced is then
λ=
(b)
yielding
h
matom λ
=
6.626 × 10 −34 J ⋅ s
= 0.814 m s
(1.67 × 10 −27 kg) ( 4.87 × 10 −7 m )
The wave function for the 2s state is given by Equation 42.26:
1 ⎛ 1⎞
ψ 2s (r ) =
4 2π ⎜⎝ a0 ⎟⎠
(a)
32
⎡
r ⎤ − r 2 a0
.
⎢2 − ⎥ e
a0 ⎦
⎣
Taking
r = a0 = 0.529 × 10 −10 m
we find
ψ 2 s ( a0 ) =
1 ⎛
1
⎞
4 2π ⎝ 0.529 × 10 −10 m ⎠
32
[ 2 − 1] e −1 2 = 1.57 × 1014 m −3 2
(b)
ψ 2 s ( a0 ) = (1.57 × 1014 m −3 2 ) = 2.47 × 10 28 m −3
(c)
Using Equation 42.24 and the results to (b) gives P2 s ( a0 ) = 4π a02 ψ 2 s ( a0 ) = 8.69 × 108 m −1 .
2
2
2
Atomic Physics
*P42.58 The average squared separation distance is
r2 = ∫
all space
We use
∫
∞
0
x n e− ax dx =
r2 =
∆r =
P42.59
πa
3
0
e −r a0 r 2
πa
3
0
e −r a0 4π r 2 dr =
4π
π a03
∫
∞
0
r 4 e −2r a0 dr
a02 96
4
4!
= 3a02
5 =
3
32
a0 ( 2 / a0 )
r
2
− r
2
2
⎛
⎛ 3a0 ⎞ ⎞
= ⎜ 3a02 −
⎝ 2 ⎠ ⎟⎠
⎝
12
⎛
9a 2 ⎞
= ⎜ 3a02 − 0 ⎟
⎝
4 ⎠
12
3
= ⎛ ⎞
⎝ 4⎠
12
a0
m s ) (14.0 × 10 −12 s ) = 4.20 mm
(b)
E=
hc
= 2.86 × 10 −19 J
λ
N=
3.00 J
= 1.05 × 1019 photons
2.86 × 10 −19 J
8
1
n!
from Table B.6.
a n+1
(3.00 × 10
2
V = ( 4.20 mm ) ⎡⎣π ( 3.00 mm ) ⎤⎦ = 119 mm 3
n=
P42.61
1
(a)
(c)
P42.60
∞
r =0
ψ 1∗s r 2ψ 1s dV = ∫
1.05 × 1019
= 8.82 × 1016 mm −3
119
(a)
The length of the pulse is ∆L = c∆t .
(b)
The energy of each photon is
(c)
V = ∆Lπ
d2
4
Eγ =
n=
hc
E
Eλ
so N =
=
Eγ
hc
λ
⎛ 4 ⎞ ⎛ Eλ ⎞
N
= ⎜
⎟
⎜
⎝ c∆tπ d 2 ⎟⎠ ⎝ hc ⎠
V
The fermions are described by the exclusion principle. Two of them, one spin-up and one
spin-down, will be in the ground energy level, with
d NN = L =
1
h
h
λ , λ = 2L = , and p =
2
p
2L
K=
1
p2
h2
=
mv 2 =
2
2m 8mL2
K=
p2
h2
=
2m 2mL2
The third must be in the next higher level, with
d NN =
L λ
h
= , λ = L, and p =
2 2
L
The total energy is then
h2
h2
h2
3h 2
+
+
=
8mL2 8mL2 2mL2
4 mL2
513
514
P42.62
Chapter 42
∆z =
2
at 2 1 ⎛ Fz ⎞ 2 µ z ( dBz dz ) ⎛ ∆ x ⎞
= ⎜ ⎟t =
⎝ v ⎠
2
2 ⎝ m0 ⎠
2 m0
and
e
2me
µz =
−27
−3
4
2
2
−31
dBz 2m0 ( ∆ z )v 2 ( 2me ) 2 (108 ) (1.66 × 10 kg ) (10 m ) (10 m s ) ( 2 × 9.11 × 10 kg )
=
=
dz
∆ x 2 e
(1.00 m 2 ) (1.60 × 10 −19 C) (1.05 × 10 −34 J ⋅ s )
dBz
= 0.389 T m
dz
P42.63
−1 2 ⎛
1
r⎞
2π a03 ) ⎜ 2 − ⎟ e − r 2 a0
(
4
a0 ⎠
⎝
We use
ψ 2s (r ) =
By Equation 42.24,
1 ⎛ r2 ⎞ ⎛
r⎞
P ( r ) = 4π r ψ = ⎜ 3 ⎟ ⎜ 2 − ⎟ e − r a0
8 ⎝ a0 ⎠ ⎝
a0 ⎠
2
(a)
2
2
2
2
r ⎞ r2 ⎛
r ⎞ ⎛ 1 ⎞⎤
2r 2 ⎛ 1 ⎞ ⎛
dP ( r ) 1 ⎡ 2r ⎛
r⎞
= ⎢ 3 ⎜ 2 − ⎟ − 3 ⎜ ⎟ ⎜ 2 − ⎟ − 3 ⎜ 2 − ⎟ ⎜ ⎟ ⎥ e − r a0 = 0
a0 ⎠ a0 ⎝
a0 ⎠ ⎝ a0 ⎠ ⎥⎦
8 ⎢⎣ a0 ⎝
dr
a0 ⎠
a0 ⎝ a0 ⎠ ⎝
or
1⎛ r ⎞ ⎛
r ⎞⎡ ⎛
r ⎞ 2r r ⎛
r ⎞⎤
2 − ⎟ ⎢ 2 ⎜ 2 − ⎟ − − ⎜ 2 − ⎟ ⎥ e − r a0 = 0
3⎟⎜
⎜
8 ⎝ a0 ⎠ ⎝
a0 ⎠ ⎣ ⎝
a0 ⎠ a0 a0 ⎝
a0 ⎠ ⎦
dP
= 0 at r = 0, r = 2a0 , and r = ∞ are minima with P ( r ) = 0.
dr
The roots of
2
⎛ 6r ⎞ ⎛ r ⎞
=0
+
⎝ a0 ⎟⎠ ⎜⎝ a0 ⎟⎠
Therefore we require
[.....] = 4 − ⎜
with solutions
r = 3 ± 5 a0
)
(
We substitute the last two roots into P ( r ) to determine the most probable value:
(
)
P (r ) =
0.051 9
a0
When r = 3 + 5 a0 = 5.236a0 ,
(
)
P (r ) =
0.191
a0
Therefore, the most probable value of r is
(3 + 5 ) a
When r = 3 − 5 a0 = 0.763 9a0 ,
= 5.236a0
2
∞
(b)
0
∞
1 ⎛ r2 ⎞ ⎛
r ⎞ −r a
)
P
r
dr
=
(
∫0
∫0 8 ⎜⎝ a03 ⎟⎠ ⎜⎝ 2 − a0 ⎟⎠ e 0 dr
Let u =
∞
r
, dr = a0 du,
a0
∞
∞
1 2
1 4
2
−u
−u
2
3
∫0 P (r ) dr = ∫0 8 u ( 4 − 4u + u ) e dr = ∫0 8 (u − 4u + 4u ) e du
∞
=−
1 4
(u + 4u 2 + 8u + 8) e− u = 1
8
0
This is as required for normalization.
Atomic Physics
P42.64
(a)
515
Suppose the atoms move in the +x direction. The absorption of a photon by an atom is a
completely inelastic collision, described by
mv i ˆi +
( )
h ˆ
− i = mv f ˆi
λ
v f − vi = −
so
h
mλ
This happens promptly every time an atom has fallen back into the ground state, so it
happens every 10 −8 s = ∆t. Then,
a=
(b)
P42.65
v f − vi
h
6.626 × 10 −34 J ⋅ s
=−
~−
~ −10 6 m s 2
mλ∆t
∆t
(10 −25 kg) (500 × 10 −9 m ) (10 −8 s )
With constant average acceleration,
v 2f = vi2 + 2a∆x
0 ~ (10 3 m s ) + 2 ( −10 6 m s 2 ) ∆x
so
∆x ~
2
(10
3
m s)
2
10 6 m s 2
~1 m
With one vacancy in the K shell, excess energy
1 1
2
∆E ≈ − ( Z − 1) (13.6 eV ) ⎛ 2 − 2 ⎞ = 5.40 keV
⎝2 1 ⎠
We suppose the outermost 4s electron is shielded by 22 electrons inside its orbit:
Eionization ≈
22 (13.6 eV )
= 3.40 eV
42
As evidence that this is of the right order of magnitude, note that the experimental ionization
energy is 6.76 eV.
K = ∆E − Eionization ≈ 5.39 keV
P42.66
E=
hc 1 240 eV ⋅ nm
=
= ∆E
λ
λ
λ1 = 310 nm,
∆E1 = 4.00 eV
so
λ2 = 400 nm,
∆E2 = 3.10 eV
λ3 = 1 378 nm,
∆E3 = 0.900 eV
FIG. P42.66
and the ionization energy = 4.10 eV.
The energy level diagram having the fewest levels and consistent with these energies is shown at
the right.
∞
P42.67
P=
∞
4r 2 −2r a0
1
2r
e
dr = ∫ z 2 e − z dz where z ≡
3
∫
2 5.00
a0
a0
2.50 a0
P=−
1 2
( z + 2z + 2) e− z
2
∞
5.00
1
1
37
= − [ 0 ] + ( 25.0 + 10.0 + 2.00 ) e −5 = ⎛ ⎞ ( 0.006 74 ) = 0.125
⎝ 2⎠
2
2
516
P42.68
Chapter 42
(a)
One molecule’s share of volume
Al:
V=
mass per molecule ⎛
27.0 g mol
⎞ ⎛ 1.00 × 10 −6 m 3 ⎞
=⎜
23
⎟⎠
⎝ 6.02 × 10 molecules mol ⎟⎠ ⎜⎝
density
2.70 g
= 1.66 × 10 −29 m 3
3
U:
−6
3
238 g
⎞ ⎛ 1.00 × 10 m ⎞ = 2.09 × 10 −29 m 3
V =⎛
23
⎟⎠
⎝ 6.02 × 10 molecules ⎠ ⎜⎝
18.9 g
3
(b)
P42.69
V = 2.55 × 10 −10 m~10 −1 nm
V = 2.76 × 10 −10 m ~ 10 −1 nm
The outermost electron in any atom sees the nuclear charge screened by all the electrons
below it. If we can visualize a single outermost electron, it moves in the electric field of
net charge + Ze − ( Z − 1) e = + e , the charge of a single proton, as felt by the electron in
hydrogen. So the Bohr radius sets the scale for the outside diameter of every atom. An
innermost electron, on the other hand, sees the nuclear charge unscreened, and the scale
a
size of its (K-shell) orbit is 0 .
Z
∆E = 2 µ B B = hf
so 2 ( 9.27 × 10 −24 J T ) ( 0.350 T ) = ( 6.626 × 10 −34 J ⋅ s ) f
and f = 9.79 × 10 9 Hz
ANSWERS TO EVEN PROBLEMS
1
1/λCA − 1/λ BA
P42.2
(a) 1.94 mm
(b) λCB =
P42.4
(a) 56.8 fm
(b) 11.3 N away from the nucleus
P42.6
(a) 2.19 Mm/s
P42.8
(a) The atoms must be excited to energy level n = 4, to emit six different photon energies in
the downward transitions 4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, and 2 → 1. The photon energy
absorbed in the 1 → 4 transition is 12.8 eV, making the wavelength 97.4 nm. (b) 1.88 mm,
infrared, Paschen (c) 97.4 nm, ultraviolet, Lyman
P42.10
(a) 3
P42.12
(a) 153 as (b) 8.18 × 109 revolutions (c) Its lifetime in electron years is comparable to the
lifetime of the Sun in Earth years, so you can think of it as a long time.
P42.14
(a) See the solution.
P42.16
See the solution.
P42.18
4 a0
P42.20
797 times
(b) 13.6 eV
(c) −27.2 eV
(b) 520 km s
(b) 0.179 nm
Atomic Physics
P42.22
6 = 2.58 × 10 −34 J ⋅ s
(a)
(b)
12 = 3.65 × 10 −34 J ⋅ s
P42.24
6
P42.26
6; −2, −, 0, , 2; 145°, 114°, 90.0°, 65.9°, 35.3°
P42.28
3
P42.30
(a) 1s 2 2 s 2 2 p 4
(b) n
m
ms
1 1
0 0
0 0
1 1
−
2 2
2 2
0 0
0 0
1 1
−
2 2
517
2 2
1 1
1 1
1 1
−
2 2
2 2
1 1
0 −1
1 1
2 2
P42.32
See the solution.
P42.34
(a) See the solution.
P42.36
1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4 d 10 4 f 14 5s 2 5 p 6 5d 10 5 f 14 6 s 2 6 p 6 6d 8 7 s 2
P42.38
(a) = 0 with m = 0; = 1 with m = 1, 0, or 1; and = 2 with m = −2, –1, 0, 1, 2 (b) –6.05 eV
P42.40
(a) ∆V = 1240 V ⋅ nm/l. (b) The potential difference is inversely proportional to the wavelength.
(c) Yes. It predicts a minimum wavelength of 33.5 pm when the accelerating voltage is 37 keV,
in agreement with the minimum wavelength in the figure. (d) Yes, but it might be unlikely for a
very high-energy electron to stop in a single interaction to produce a high-energy gamma ray; and
it might be difficult to observe the very low-intensity radio waves produced as bremsstrahlung
by low-energy electrons. The potential difference goes to infinity as the wavelength goes to zero.
The potential difference goes to zero as the wavelength goes to infinity.
P42.42
L shell 11.8 keV, M shell 10.1 keV, N shell 2.39 keV. See the solution.
P42.44
See the solution.
P42.46
(a) 4.24 PW m 2
P42.48
(a) 1.07 × 10 −33 (b) −1.15 × 10 6 K
thermal equilibrium.
P42.50
(a) See the solution.
P42.52
⎛ 2r 2 2r ⎞
(a) Probability = ⎜ 2 + + 1⎟ e −2r a0
⎝ a0 a0 ⎠
P42.54
(a) 10.2 eV = 1.63 aJ
P42.56
(a) 487 nm
P42.58
(b) 36 states instead of 30
(b) 1.20 pJ = 7.50 MeV
(b) 2.53 × 10 74
(c) 1.18 × 10 −63 m , unobservably small
(b) 7.88 × 10 4 K
(b) 0.814 m s
3/4 a0 = 0.866 a0
(c) Negative temperatures do not describe systems in
(b) See the solution.
(c) 1.34 a0
518
Chapter 42
Eλ
hc
4 Eλ
∆tπ d 2 hc 2
P42.60
(a) c∆t
P42.62
0.389 T/m
P42.64
(a) ~ –106 m/s2 (b) ~1 m
P42.66
Energy levels at 0, –0.100 eV, –1.00 eV, and –4.10 eV
P42.68
(a) diameter ~10 −1 nm for both
order of
(b)
a0
.
Z
(c)
(b) A K-shell electron moves in an orbit with size on the
43
Molecules and Solids
CHAPTER OUTLINE
43.1
43.2
43.3
43.4
43.5
43.6
43.7
43.8
Molecular Bonds
Energy States and Spectra of
Molecules
Bonding in Solids
Free-Electron Theory of Metals
Band Theory of Solids
Electrical Conduction in Metals,
Insulators, and Semiconductors
Semiconductor Devices
Superconductivity
Q43.2
ANSWERS TO QUESTIONS
Q43.1
Ionic bonds are ones between oppositely charged ions.
A simple model of an ionic bond is the electrostatic
attraction of a negatively charged latex balloon to a
positively charged Mylar balloon.
Covalent bonds are ones in which atoms share
electrons. Classically, two children playing a short-range
game of catch with a ball models a covalent bond. On
a quantum scale, the two atoms are sharing a wave
function, so perhaps a better model would be two
children using a single hula hoop.
Van der Waals bonds are weak electrostatic forces:
the dipole-dipole force is analogous to the attraction
between the opposite poles of two bar magnets, the
dipole-induced dipole force is similar to a bar magnet
attracting an iron nail or paper clip, and the dispersion
force is analogous to an alternating-current electromagnet attracting a paper clip.
A hydrogen atom in a molecule is not ionized, but its
electron can spend more time elsewhere than it does in
the hydrogen atom. The hydrogen atom can be a location
of net positive charge, and can weakly attract a zone of
negative charge in another molecule.
Rotational, vibrational, and electronic (as discussed in Chapter 42) are the three major forms
2
of excitation. Rotational energy for a diatomic molecule is on the order of
, where I is
2I
the moment of inertia of the molecule. A typical value for a small molecule is on the order of
1 meV = 10 −3 eV. Vibrational energy is on the order of hf, where f is the vibration frequency
of the molecule. A typical value is on the order of 0.1 eV. Electronic energy depends on the
state of an electron in the molecule and is on the order of a few eV. The rotational energy can
be zero, but neither the vibrational nor the electronic energy can be zero.
*Q43.3 If you start with a solid sample and raise its temperature, it will typically melt first, then start
emitting lots of far infrared light, then emit light with a spectrum peaking in the near infrared,
and later have its molecules dissociate into atoms. Rotation of a diatomic molecule involves less
energy than vibration. Absorption and emission of microwave photons, of frequency ~1011 Hz,
accompany excitation and de-excitation of rotational motion, while infrared photons, of
frequency ~1013 Hz, accompany changes in the vibration state of typical simple molecules.
The ranking is then b > d > c > a.
519
520
Q43.4
Chapter 43
From the rotational spectrum of a molecule, one can easily calculate the moment of inertia of
the molecule using Equation 43.7 in the text. Note that with this method, only the spacing
between adjacent energy levels needs to be measured. From the moment of inertia, the size of the
molecule can be calculated, provided that the structure of the molecule is known.
*Q43.5 Answer (b). At higher temperature, molecules are typically in higher rotational energy levels
before as well as after infrared absorption.
*Q43.6 (i) Answer (a). An example is NaCl, table salt.
(ii) Answer (b). Examples are elemental silicon and carborundum (silicon carbide).
(iii) Answer (c). Think of aluminum foil.
*Q43.7 (i) Answer (b). The density of states is proportional to the energy to the one-half power.
(ii) Answer (a). Most states well above the Fermi energy are unoccupied.
Q43.8
In a metal, there is no energy gap between the valence and conduction bands, or the conduction
band is partly full even at absolute zero in temperature. Thus an applied electric field is able to
inject a tiny bit of energy into an electron to promote it to a state in which it is moving through
the metal as part of an electric current. In an insulator, there is a large energy gap between a full
valence band and an empty conduction band. An applied electric field is unable to give electrons
in the valence band enough energy to jump across the gap into the higher energy conduction
band. In a semiconductor, the energy gap between valence and conduction bands is smaller than
in an insulator. At absolute zero the valence band is full and the conduction band is empty, but
at room temperature thermal energy has promoted some electrons across the gap. Then there are
some mobile holes in the valence band as well as some mobile electrons in the conduction band.
*Q43.9 Answer (b). First consider electric conduction in a metal. The number of conduction electrons
is essentially fixed. They conduct electricity by having drift motion in an applied electric field
superposed on their random thermal motion. At higher temperature, the ion cores vibrate more
and scatter more efficiently the conduction electrons flying among them. The mean time between
collisions is reduced. The electrons have time to develop only a lower drift speed. The electric
current is reduced, so we see the resistivity increasing with temperature.
Now consider an intrinsic semiconductor. At absolute zero its valence band is full and its
conduction band is empty. It is an insulator, with very high resistivity. As the temperature
increases, more electrons are promoted to the conduction band, leaving holes in the valence
band. Then both electrons and holes move in response to an applied electric field. Thus we see
the resistivity decreasing as temperature goes up.
Q43.10 The energy of the photon is given to the electron. The energy of a photon of visible light is
sufficient to promote the electron from the lower-energy valence band to the higher-energy
conduction band. This results in the additional electron in the conduction band and an additional
hole—the energy state that the electron used to occupy—in the valence band.
Q43.11 Along with arsenic (As), any other element in group V, such as phosphorus (P), antimony (Sb),
and bismuth (Bi), would make good donor atoms. Each has 5 valence electrons. Any element in
group III would make good acceptor atoms, such as boron (B), aluminum, (Al), gallium (Ga), and
indium (In). They all have only 3 valence electrons.
Q43.12 The two assumptions in the free-electron theory are that the conduction electrons are not bound to
any particular atom, and that the nuclei of the atoms are fixed in a lattice structure. In this model,
it is the “soup” of free electrons that are conducted through metals. The energy band model is
more comprehensive than the free-electron theory. The energy band model includes an account
of the more tightly bound electrons as well as the conduction electrons. It can be developed into a
theory of the structure of the crystal and its mechanical and thermal properties.
Molecules and Solids
521
Q43.13 A molecule containing two atoms of 2 H, deuterium, has twice the mass of a molecule
containing two atoms of ordinary hydrogen 1 H. The atoms have the same electronic structure, so
the molecules have the same interatomic spacing, and the same spring constant. Then the moment
of inertia of the double-deuteron is twice as large and the rotational energies one-half as large
1
as for ordinary hydrogen. Each vibrational energy level for D 2 is
times that of H 2 .
2
Q43.14 Yes. A material can absorb a photon of energy greater than the energy gap, as an electron jumps
into a higher energy state. If the photon does not have enough energy to raise the energy of the
electron by the energy gap, then the photon will not be absorbed.
*Q43.15 (i) and (ii) Answer (a) for both. Either kind of doping contributes more mobile charge carriers,
either holes or electrons.
*Q43.16 (a) false (b) false (c) true (d) true (e) true
SOLUTIONS TO PROBLEMS
Section 43.1
Molecular Bonds
(a)
(1.60 × 10 −19 ) (8.99 × 109 ) N = 0.921 × 10 −9 N
q2
F=
=
2
4π ∈0 r
(5.00 × 10 −10 )2
(b)
U=
2
P43.1
toward the other ion.
(1.60 × 10 −19 ) (8.99 × 109 ) J ≈ −2.88 eV
−q2
=−
4π ∈0 r
5.00 × 10 −10
2
P43.2
We are told
K + Cl + 0.7 eV → K + + Cl −
and
Cl + e − → Cl − + 3.6 eV
or
Cl− → Cl + e− − 3.6 eV
By substitution,
K + Cl + 0.7 eV → K + + Cl + e− − 3.6 eV
K + 4.3 eV → K + + e−
or the ionization energy of potassium is 4.3 eV .
P43.3
(a)
Minimum energy of the molecule is found from
⎡2A⎤
dU
= −12 Ar −13 + 6 Br −7 = 0 yielding r0 = ⎢ ⎥
⎣B⎦
dr
16
(b)
2
⎡ A
B2
B ⎤
⎡ 1 1⎤ B
=
−
−
=
E = U r = ⬁ − U r = r0 = 0 − ⎢ 2 2 −
⎢⎣ 4 2 ⎥⎦ A
2 A B ⎥⎦
4A
⎣ 4A B
This is also the equal to the binding energy, the amount of energy given up by the two
atoms as they come together to form a molecule.
⎡ 2 ( 0.124 × 10 −120 eV ⋅ m12 ) ⎤
r0 = ⎢
⎥
−60
6
⎢⎣ 1.488 × 10 eV ⋅ m
⎥⎦
16
(c)
(1.488 × 10
E=
4 ( 0.124 × 10
−60
eV ⋅ m 6 )
−120
= 7.42 × 10 −11 m = 74.2 pm
2
eV ⋅ m12 )
= 4.46 eV
522
P43.4
Chapter 43
(a)
We add the reactions K + 4.34 eV → K + + e −
and
I + e − → I − + 3.06 eV
to obtain
K + I → K + + I− + ( 4.34 − 3.06 ) eV
The activation energy is 1.28 eV .
(b)
13
7
dU 4 ∈ ⎡
⎛s ⎞
⎛s ⎞ ⎤
=
⎢ −12 ⎜ ⎟ + 6 ⎜ ⎟ ⎥
dr
s ⎣
⎝r⎠
⎝r⎠ ⎦
At r = r0 we have
s
= 2 −1 6
r0
13
7
dU
= 0. Here ⎛⎜s ⎞⎟ = 1 ⎛⎜s ⎞⎟
dr
2 ⎝ r0 ⎠
⎝ r0 ⎠
s = 2 −1 6 ( 0.305 ) nm = 0.272 nm = s
Then also
⎡⎛ −1 6 ⎞12 ⎛ −1 6 ⎞6 ⎤
⎡1 1⎤
2 r0
2 r0 ⎥
U ( r0 ) = 4 ∈ ⎢⎜
⎟ −⎜
⎟ + Ea = 4 ∈ ⎢ − ⎥ + Ea = − ∈ + Ea
⎣4 2⎦
⎣⎢⎝ r0 ⎠ ⎝ r0 ⎠ ⎥⎦
∈= Ea − U ( r0 ) = 1.28 eV + 3.37 eV= 4.65 eV = ∈
(c)
F (r ) = −
13
7
dU 4 ∈ ⎡ ⎛s ⎞
⎛s ⎞ ⎤
=
⎢ 12 ⎜ ⎟ − 6 ⎜ ⎟ ⎥
s ⎣ ⎝r⎠
dr
⎝r⎠ ⎦
To find the maximum force we calculate
s
rrupture
Fmax =
⎛ 42 ⎞⎟
=⎜
⎝156 ⎠
14
8
dF 4 ∈ ⎡
⎛s ⎞
⎛s ⎞ ⎤
= 2 ⎢ −156 ⎜ ⎟ + 42 ⎜ ⎟ ⎥ = 0 when
dr s ⎣
⎝r⎠
⎝r⎠ ⎦
16
13 6
76
4 ( 4.65 eV ) ⎡ ⎛ 42 ⎞
1.6 × 10 −19 Nm
⎛ 42 ⎞ ⎤
−
=
−
.
eV
nm
=
−
.
41
0
41
0
12
6
⎟
⎜⎝
⎟ ⎥
⎢ ⎜
10 −9 m
156 ⎠ ⎦
0.272 nm ⎣ ⎝ 156 ⎠
= −6.55 nN
Therefore the applied force required to rupture the molecule is +6.55 nN away from the
center.
⎡ ⎛ 2 −1 6 r ⎞ 12 ⎛ 2 −1 6 ⎞ 6 ⎤
⎤
0
⎥ + Ea = 4 ∈ ⎢ ⎜
⎟ −⎜
⎟ ⎥ + Ea
⎥⎦
⎢⎣ ⎝ r0 + s ⎠
⎝ r0 + s ⎠ ⎥⎦
−12
−6 ⎤
⎡ ⎛
s⎞
1
1⎛ s ⎞ ⎥
⎢
= 4 ∈ ⎜1 + ⎟ − ⎜1 + ⎟ + Ea
2 ⎝ r0 ⎠ ⎥⎦
⎢⎣ 4 ⎝ r0 ⎠
⎞ 1⎛
⎞⎤
⎡1 ⎛
s
s2
s
s2
= 4 ∈ ⎢ ⎜1 − 12 + 78 2 −⎟ − ⎜1 − 6 + 21 2 −⎟⎥ + Ea
r0
r0
r0
r0
⎠ 2⎝
⎠⎦
⎣4 ⎝
⎡⎛ s ⎞
⎛ s ⎞
U ( r0 + s ) = 4 ∈ ⎢ ⎜
⎟ −⎜
⎟
⎢⎣ ⎝ r0 + s ⎠
⎝ r0 + s ⎠
12
(d)
6
s
s2
s
s2
+ 78 ∈ 2 − 2 ∈ +12 ∈ − 42 ∈ 2 + Ea + r0
r0
r0
r0
2
⎛s⎞
s
= − ∈ +Ea + 0 ⎜ ⎟ + 36 ∈ 2 + r
r
⎝ 0⎠
0
1 2
U ( r0 + s ) ≈ U ( r0 ) + ks
2
= ∈ −12 ∈
where k =
72 ∈ 72 ( 4.65 eV)
=
= 3 599 eV nm 2 = 576 N m
2
r02
(0.305 nm )
Molecules and Solids
P43.5
At the boiling or condensation temperature,
kBT ≈ 10 −3 eV = 10 −3 (1.6 × 10 −19 J )
T≈
Section 43.2
*P43.6
(a)
523
1.6 × 10 −22 J
~ 10 K
1.38 × 10 −23 J K
Energy States and Spectra of Molecules
With r representing the distance of each atom from the center of mass, the moment of
inertia is
mr + mr = 2 (1.67 × 10
2
2
2
−27
⎛ 0.75 × 10 −10 m ⎞
−48
2
kg ) ⎜
⎟⎠ = 4.70 × 10 kg ⋅ m
⎝
2
The rotational energy is Erot =
2
J ( J + 1) or it is zero for J = 0 and for J = 1 it is
2I
(6.626 × 10−34 J ⋅ s) ⎛⎜ 1 eV ⎞⎟ = 0.0148 eV
h 2 1(2 )
=
4 π 2 2 I 4 π 2 ( 4.70 × 10 −48 kg ⋅ m 2 ) ⎝ 1.6 × 10 −19 J ⎠
2
(b)
P43.7
µ=
λ=
c ch (2.998 × 10 8 m/s)(6.626 × 10 −34 J ⋅ s)
1 eV
1240 eV ⋅ nm
=
= 83.8 µ m
=
=
−19
f E
0.00148 eV
1.602 × 10 J
0.0148 eV
m1 m2
132.9 (126.9 )
=
(1.66 × 10 −27 kg ) = 1.08 × 10 −25 kg
m1 + m2 132.9 + 126.9
I = µ r 2 = (1.08 × 10 −25 kg ) ( 0.127 × 10 −9 m ) = 1.74 × 10 −45 kg ⋅ m 2
2
(a)
1 2 ( Iω )2 J ( J + 1) 2
Iω =
=
2
2I
2I
J = 0 gives E = 0
E=
(6.626 × 10 −34 J ⋅ s) = 6.41 × 10 −24 J = 40.0 µeV
2
=
I
4π 2 (1.74 × 10 −45 kg ⋅ m 2 )
2
J = 1 gives E =
hf = 6.41 × 10 −24 J − 0 gives f = 9.66 × 10 9 Hz
(b)
*P43.8
f=
∆Evib =
E1 2
h
=
=
h hI 4 π 2 µ r 2
h
2π
k
= hf
µ
∝ r −2
so
If r is 10% too small, f is 20% too large.
k = 4π 2 f 2 µ
µ=
k
1530 N/m
=
= 1.22 × 10 −26 kg
4π 2 f 2 4π 2 (56.3 × 1012 /s)2
µ=
14.007 u 15.999 u 1.66 × 10 −27 kg
m1 m2
= 1.24 × 10 −26 kg
=
1u
m1 + m2 14.007 u +15.9999 u
The reduced masses agree, because the small apparent difference can be attributed to uncertainty
in the data.
524
P43.9
Chapter 43
For the HCl molecule in the J = 1 rotational energy level,
we are given r0 = 0.127 5 nm.
Taking J = 1 , we have
Erot =
2
J ( J + 1)
2I
Erot =
2 1 2
22
= Iω or ω =
= 2
2
2
I
I
I
The moment of inertia of the molecule is given by Equation 43.3.
FIG. P43.9
⎛ mm ⎞
I = µ r02 = ⎜ 1 2 ⎟ r02
⎝ m1 + m2 ⎠
⎡ (1.01 u )( 35.5 u ) ⎤ 2
2
−27
−10
−47
2
I =⎢
⎥ r0 = ( 0.982 u ) (1.66 × 10 kg u ) (1.275 × 10 m ) = 2.665 × 10 kg ⋅ m
u
u
1
01
35
5
.
.
+
⎣
⎦
Therefore, ω = 2
P43.10
hf = ∆E =
2 6.626 × 10 −34 J ⋅ s
=
= 5.63 × 1012 rad s
I 2π (2.65 × 10 −47 kg ⋅ m 2 )
2
2
2
2 ( 2 + 1)] − [1(1 + 1)] = ( 4 )
[
2I
2I
2I
4 ( h 2π )
h
6.626 × 10 −34 J ⋅ s
=
=
= 1.46 × 10 −46 kg ⋅ m 2
2
2 hf
2π f 2π 2 ( 2.30 × 1011 Hz )
2
I=
P43.11
I = m1r12 + m2 r22
mr
Then r1 = 2 2
m1
m1r1
Also, r2 =
.
m2
I = m1
P43.12
(a)
where
so
Thus,
m22 r 2
( m1 + m2 )
µ=
2
+
m2 m12 r 2
( m1 + m2 )
2
=
m1r1 = m2 r2
m2 r2
+ r2 = r
m1
mr
r1 + 1 1 = r
m2
m1 m2 r 2 ( m2 + m1 )
( m1 + m2 )
2
=
and
and
and
r1 + r2 = r
m1r
r2 =
m1 + m2
m2 r
r1 =
m1 + m2
m1 m2 r 2
= µr 2
m1 + m2
22.99 (35.45)
(1.66 × 10 −27 kg ) = 2.32 × 10−26 kg
( 22.99 + 35.45)
I = µ r 2 = ( 2.32 × 10 −26 kg ) ( 0.280 × 10 −9 m ) = 1.81 × 10 −45 kg ⭈ m 2
2
(b)
hc 2
2
3 2 2 2 2
2h2
=
−
=
=
2 ( 2 + 1) − 1(1 + 1) =
λ 2I
2I
I
I
I
4π 2 I
λ=
−45
2
8
2
c 4π 2 I (3.00 × 10 m s ) 4π (1.81 × 10 kg ⋅ m )
=
= 1.62 cm
2h
2 (6.626 × 10 −34 J ⋅ s )
Molecules and Solids
P43.13
⎛ 2 ⎞
The energy of a rotational transition is ∆E = ⎜ ⎟ J where J is the rotational quantum number of
⎝ I ⎠
the higher energy state (see Equation 43.7). We do not know J from the data. However,
∆E =
hc (6.626 × 10
=
λ
−34
J ⋅ s ) (3.00 × 108 m s ) ⎛
1 eV
⎞
⎜⎝
⎟.
λ
1..60 × 10 −19 J ⎠
For each observed wavelength,
l (mm)
∆E (eV)
0.120 4
0.010 32
0.096 4
0.012 88
0.080 4
0.015 44
0.069 0
0.018 00
0.060 4
0.020 56
E1 =
The ∆E⬘s consistently increase by 0.002 56 eV.
2
= 0.002 56 eV
I
−34
⎞
1 eV
2 (1.055 × 10 J ⋅ s) ⎛
−47
2
=
⎜
⎟ = 2.72 × 10 kg ⋅ m
−19
E1
(0.002 56 eV) ⎝ 1.660 × 10 J ⎠
2
and I =
r=
For the HCl molecule, the internuclear radius is
*P43.14 (a)
2.72 × 10 −47
I
m = 0.130 nm
=
µ
1.62 × 10 −27
Minimum amplitude of vibration of HI is characterized by
1 2 1
h
kA = ω =
2
2
4π
k
µ
so
A=
h ⎛ 1 ⎞
2π ⎜⎝ k µ ⎟⎠
1/ 4
⎞
6.626 × 10 −34 J ⋅ s ⎛
1
A=
−27
⎜
⎝ (320 N/m)(127/128)(1.66 × 10 kg) ⎟⎠
2π
P43.15
525
For HF, A =
(c)
Since HI has the smaller k, it is more weakly bound.
m1 m2
35
=
× 1.66 × 10 −27 kg = 1.61 × 10 −27 kgg
m1 + m2 36
∆Evib = = 12.1 pm
⎞
6.626 × 10 −34 J ⋅ s ⎛
1
−27
⎜
⎝ ( 970 N/m)(19/20)(1.66 × 10 kg) ⎟⎠
2π
(b)
µ=
1/ 4
k
480
= 5.74 × 10 −20 J = 0.358 eV
= (1.055 × 10 −34 )
µ
1.61 × 10 −27
1/ 4
= 9.23 pm
526
P43.16
Chapter 43
(a)
The reduced mass of the O 2 is
µ=
(16 u)(16 u) = 8 u = 8 1.66 × 10−27
(
(16 u) + (16 u)
kg ) = 1.33 × 10 −26 kg
The moment of inertia is then
I = µ r 2 = (1.33 × 10 −26 kg ) (1.20 × 10 −10 m )
2
= 1.91 × 10 −46 kg ⋅ m 2
(1.055 × 10−34 J ⋅ s) J ( J + 1)
2
The rotational energies are Erot =
J ( J + 1) =
2I
2 (1.91 × 10 −46 kg ⋅ m 2 )
2
(b)
Thus
Erot = ( 2.91 × 10 −23 J ) J ( J + 1)
And for J = 0, 1, 2,
Erot = 0, 3.64 × 10 −4 eV, 1.09 × 10 −3 eV
⎛
1177 N m
1⎞ k ⎛
1⎞
Evib = ⎜ v + ⎟ = ⎜ v + ⎟ (1.055 × 10 −34 J ⋅ s)
⎝
2⎠ µ ⎝
2⎠
8 (1.66 × 10 −27 kg )
1⎞
1⎞
1 eV
⎛
⎞⎟ ⎛⎜
⎛
Evib = ⎜v + ⎟ (3.14 × 10 −20 J ) ⎜
= v + ⎟ ( 0.196 eV )
−19
2⎠
2⎠
⎝1.60 × 10 J ⎠ ⎝
⎝
For v = 0, 1, 2,
P43.17
Evib = 0.098 2 eV, 0.295 eV, 0.491 eV .
In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is
( 0.110 + 0.100 nm ) = 0.210 nm from the axis. Thus, I = Σmr 2 :
I = 6 (1.99 × 10 −26 kg ) ( 0.110 × 10 −9 m ) + 6 (1.67 × 10 −27 kg ) ( 0.210 × 10 −9 m )
2
2
= 1.89 × 10 −45 kg ⋅ m 2
The allowed rotational energies are then
(1.055 × 10 −34 J ⋅ s ) J ( J + 1) = 2.95 × 10 −24 J J ( J + 1)
2
=
J ( J + 1) =
(
)
2I
2 (1.89 × 10 −45 kg ⋅ m 2 )
2
Erot
= (18.4 × 10 −6 eV ) J ( J + 1)
Erot = (18.4 µ eV) J ( J + 1) where J = 0, 1, 2, 3, . . .
The first five of these allowed energies are: Erot = 0, 36.9 µ eV, 111 µ eV, 221 µ eV, and 369 µ eV.
Molecules and Solids
P43.18
527
We carry extra digits through the solution because part (c) involves the subtraction of two close
numbers. The longest wavelength corresponds to the smallest energy difference between the
2
rotational energy levels. It is between J = 0 and J = 1, namely
I
4π 2 Ic
hc
hc
λ=
=
=
. If m is the reduced mass, then
∆Emin 2 I
h
I = µ r 2 = µ ( 0.127 46 × 10 −9 m ) = (1.624 605 × 10 −20 m 2 ) µ
2
Therefore λ =
µ35 =
(a)
4 π 2 (1.624 605 × 10 −20 m 2 ) µ ( 2.997 925 × 10 8 m s)
6.626 075 × 10 −34 J ⋅ s
(1)
= ( 2.901 830 × 10 23 m kg ) µ
(1.007 825u )(34.968 853u ) = 0.979 593u = 1.626 653 × 10 −27
1.007 825u + 34.968 853u
kg
From (1): λ35 = ( 2.901 830 × 10 23 m kg ) (1.626 653 × 10 −27 kg ) = 472 µ m
µ37 =
(b)
(1.007 825u )(36.965 903u ) = 0.981 077u = 1.629 118 × 10 −27
1.007 825u + 36.965 903u
kg
From (1): λ37 = ( 2.901 830 × 10 23 m kg ) (1.629 118 × 10 −27 kg ) = 473 µ m
λ37 − λ35 = 472.742 4 µ m − 472.027 0 µ m = 0.715 µ m
(c)
P43.19
We find an average spacing between peaks by counting 22 gaps between 7.96 × 1013 Hz and
9.24 × 1013 Hz:
∆f =
I=
(9.24 − 7.96)1013 Hz = 0.058 2 × 1013 Hz = 1 ⎛
22
h2 ⎞
⎜ 2 ⎟
h ⎝ 4π I ⎠
6.63 × 10 −34 J ⋅ s
h
= 2.9 × 10 −47 kg ⋅ m 2
=
4 π 2 ∆f 4 π 2 5.82 × 1011 s
528
P43.20
Chapter 43
We carry extra digits through the solution because the given wavelengths are close together.
(a)
1⎞
2
⎛
EvJ = ⎜v + ⎟ hf +
J ( J + 1)
2⎠
2I
⎝
∴ E00 =
1
hf ,
2
E11 =
∴ E11 − E00 = hf +
3
2
1
3 2
hf + , E02 = hf +
I
I
2
2
−34
8
2 hc (6.626 075 × 10 J ⋅ s ) ( 2.997 925 × 10 m s )
=
=
I
λ
2.211 2 × 10 −6 m
2
= 8.983 573 × 10 −20 J
I
−34
8
2 2 hc (6.626 075 × 10 J ⋅ s ) ( 2.997 925 × 10 m s )
=
=
E11 − E02 = hf −
I
λ
2.405 4 × 10 −6 m
2
2
∴ hf −
= 8.258 284 × 10 −20 J
I
3 2
= 7.252 89 × 10 −21 J
Subtract (2) from (1):
I
2
3 (1.054 573 × 10 −34 J ⋅ s )
∴I =
= 4.60 × 10 −48 kg ⋅ m 2
7.252 89 × 10 −21 J
∴ hf +
(b)
(1)
(2)
From (1):
(1.054 573 × 10 −34 J ⭈ s)
8.983 573 × 10 −20 J
−
6.626 075 × 10 −34 J ⭈ s ( 4.600 060 × 10 −48 kg ⭈ m 2 ) (6.626 075 × 10 −34 J ⭈ s )
2
f=
= 1.32 × 1014 Hz
(c)
I = µ r 2, where m is the reduced mass:
1
1
µ = mH = (1.007 825 u ) = 8.367 669 × 10 −28 kg
2
2
So r =
P43.21
I
=
µ
4.600 060 × 10 −48 kg ⋅ m 2
= 0.074 1 nm .
8.367 669 × 10 −28 kg
The emission energies are the same as the absorption energies, but the final state must be below
( v = 1, J = 0 ) . The transition must satisfy ∆J = ±1, so it must end with J = 1. To be lower in
energy, it must be ( v = 0, J = 1) . The emitted photon energy is therefore
(
hfphoton = Evib v=1 + Erot
J =0
) − (E
vib v = 0
+ Erot
J =1
) = (E
vib v =1
) (
− Evib v= 0 − Erot
J =1
− Erot
J =0
)
hfphoton = hfvib − hfrot
Thus, fphoton = fvib − frot = 6.42 × 1013 Hz − 1.15 × 1011 Hz = 6.41 × 1013 Hz .
P43.22
The moment of inertia about the molecular axis is I x =
2
2 2 2 2 4
mr + mr = m ( 2.00 × 10 −15 m ) .
5
5
5
2
⎛R⎞
⎛R⎞ m
The moment of inertia about a perpendicular axis is I y = m ⎜ ⎟ + m ⎜ ⎟ = ( 2.00 × 10 −10 m ) .
⎝2⎠
⎝2⎠
2
2
2
⎛ 2 ⎞
The allowed rotational energies are Erot = ⎜ ⎟ J ( J + 1) , so the energy of the first excited state is
⎝ 2I ⎠
E1 =
E1, x
E1, y
2
. The ratio is therefore
I
(
=
(
2
2
Ix )
Iy )
=
Iy
Ix
(1 2) m (2.00 × 10−10 m )
=
2
( 4 5) m (2.00 × 10−15 m )
2
=
2
5
10 5 ) = 6.25 × 10 9
(
8
Molecules and Solids
Section 43.3
529
Bonding in Solids
P43.23
(1.60 × 10−19 ) ⎛⎜1 − 1 ⎞⎟ = −1.25 × 10−18 J = −7.84 eV
α k e2 ⎛ 1 ⎞
U = − e ⎜1 − ⎟ = − (1.747 6 ) ( 8.99 × 10 9 )
r0 ⎝ m ⎠
(0.281 × 10−9 ) ⎝ 8 ⎠
P43.24
Consider a cubical salt crystal of edge length 0.1 mm.
2
The number of atoms is
⎛
⎞
10 −4 m
⎜
⎟
−9
⎝ 0.261 × 10 m ⎠
This number of salt crystals would have volume
(10
−4
3
~ 1017
⎞
3⎛
10 −4 m
m) ⎜
⎟
−9
⎝ 0.261 × 10 m ⎠
3
~ 10 5 m 3
If it is cubic, it has edge length 40 m.
P43.25
P43.26
–e
–
e
+
–e
–
e
+
r
2r
3r
FIG. P43.25
Visualize a K + ion at the center of each shaded cube, a Cl − ion
at the center of each white one.
2 ( 0.314 nm ) = 0.444 nm
The distance ab is
2 ( 0.314 nm ) = 0.628 nm
Distance ac is
22 +
Distance ad is
( 2)
2
(0.314 nm ) =
Section 43.4
Free-Electron Theory of Metals
Section 43.5
Band Theory of Solids
P43.27
e
+
ke e 2 ke e 2 ke e 2 ke e 2 ke e 2 ke e 2 ke e 2 ke e 2
+
−
−
+
+
−
−
+
2r
2r
3r
3r
4r
4r
r
r
⎞
2 k e2 ⎛ 1 1 1
= − e ⎜1 − + − +⎟
⎝
⎠
2 3 4
r
2
3
x
x
x4
But, ln (1 + x ) = 1 − + − +
2
3 4
2 ke e 2
e2
so, U = −
ln 2, or U = −keα
where α = 2 ln 2
r
r
U =−
The density of conduction electrons n is given by EF =
or
8 π ⎛ 2 mEF ⎞
ne =
⎜
⎟
3 ⎝ h2 ⎠
FIG. P43.26
0.769 nm
h 2 ⎛ 3ne ⎞
⎜
⎟
2 m ⎝ 8π ⎠
23
⎡2 9.11 × 10 −31 kg ) ( 5.48 ) (1.60 × 10 −19 J )⎤
⎦
8π ⎣ (
=
3
−34
3
(6.626 × 10 J ⋅ s)
32
32
= 5.80 × 10 28 m −3
The number-density of silver atoms is
⎞
⎛ 1 atom ⎞ ⎛
1u
28
−3
nAg = (10.6 × 10 3 kg m 3 ) ⎜
⎟⎜
⎟ = 5.91 × 10 m
−27
⎝ 108 u ⎠ ⎝ 1.66 × 10 kg ⎠
So an average atom contributes
5.80
= 0.981 electron to the conduction baand .
5.91
–e
–
530
Chapter 43
*P43.28 (a)
(b)
The Fermi energy is proportional to the spatial concentration of free electrons to the
two-thirds power.
h 2 ⎛ 3ne ⎞
⎜
⎟
2 m ⎝ 8π ⎠
EF =
(6.626 × 10 J ⋅ s)
2 ( 9.11 × 10 kg ) (1.60 × 10
−34
23
=
2
−31
−19
⎛ 3⎞
⎜ ⎟
J eV ) ⎝ 8π ⎠
23
ne2 3 becomes
EF = (3.65 × 10 −19 ) ne2 3 eV with n measured in electrons m 3.
(c)
P43.29
(a)
Copper has the greater concentration of free electrons by a factor of
8.49 × 1028/1.40 × 1028 = 6.06
Copper has the greater Fermi energy by a factor of 7.05 eV/2.12 eV = 3.33. This behavior
agrees with the proportionality because 6.062/3 = 3.33.
1
mv 2 = 7.05 eV
2
2 ( 7.05 eV ) (1.60 × 10 −19 J eV )
v=
= 1.57 × 10 6 m s
9.11 × 10 −31 kg
Larger than 10 −4 m s by ten orders of magnitude. However, the energy of an electron at
(b)
room temperature is typically kBT =
1
eV.
40
*P43.30 The occupation probability is
f (E ) =
=
*P43.31 (a)
(b)
1
exp[−0.01( 7.05 eV)(1.6 × 10
Eav =
−19
1
1
= −2.72
= 0.938
−23
J/eV)/(1.38 × 10 J/K)300 K] + 1 e
+1
3
EF = 0.6( 7.05 eV) = 4.23 eV
5
The average energy of a molecule in an ideal gas is
T=
P43.32
1
=
e( E −EF )/ kBT + 1 e( 0.99 EF −EF )/ kBT + 1
3
kBT so we have
2
2
4.23 eV
1.6 × 10 −19 J
= 3.27 ×10 4 K
3 1.38 × 10 −23 J/K
1 eV
For sodium, M = 23.0 g mol and ρ = 0.971 g cm 3.
(a)
ne =
3
23
N A ρ ( 6.02 × 10 electrons mol)( 0.971 g cm )
=
23.0 g mol
M
ne = 2.54 × 10 22 electrons cm 3 = 2.54 × 10 28 electrons m 3
(b)
⎛ h 2 ⎞ ⎛ 3ne ⎞
EF = ⎜
⎝ 2 m ⎟⎠ ⎝ 8π ⎠
23
(6.626 × 10
=
2 ( 9.11 × 10
J ⋅ s ) ⎡ 3 ( 2.54 × 10 28 m −3 ) ⎤
⎢
⎥
−31
8π
kg ) ⎢⎣
⎥⎦
−34
2
= 5.05 × 10 −19 J = 3.15 eV
P43.33
Taking EF = 5.48 eV for sodium at 800 K,
−1
E −E k T
f = ⎡⎣e( F ) B + 1⎤⎦ = 0.950
1
E −E k T
e( F ) B =
− 1 = 0.052 6
0.950
E − EF
= ln ( 0.052 6 ) = −2.94
kBT
(1.38 × 10 ) (800) J = −0.203 eV or
−23
E − EF = −2.94
1.60 × 10 −19 J eV
E = 5.28 eV
23
Molecules and Solids
P43.34
531
The melting point of silver is 1 234 K. Its Fermi energy at 300 K is 5.48 eV. The approximate
fraction of electrons excited is
−23
kBT (1.38 × 10 J K ) (1 234 K )
≈ 2%
=
EF ( 5.48 eV) (1.60 × 10 −19 J eV)
P43.35
Eav =
1
ne
∞
∫ EN ( E ) dE
0
N ( E ) = 0 for E > EF
At T = 0,
Since f ( E ) = 1 for E < EEF and f ( E ) = 0 for E > EF ,
N ( E ) = CE 1 2 =
we can take
EF
EF
1
Eav =
ne
C
∫ CE dE = n
e
0
32
∫E
32
dE =
0
But from Equation 43.25,
P43.36
8 2 π me3 2 1 2
E
h3
2C 5 2
EF
5 ne
⎛ 2 ⎞⎛ 3
⎞
3
Eav = ⎜ ⎟ ⎜ EF−3 2 ⎟ EF5 2 =
EF
⎝ 5 ⎠⎝ 2
⎠
5
C 3 −3 2
= EF , so that
ne 2
d = 1.00 mm, so
V = (1.00 × 10 −3 m ) = 1.00 × 10 −9 m 3
The density of states is
g ( E ) = CE 1 2 =
or
g (E) =
3
8 2π m 3 2 1 2
E
h3
8 2π ( 9.11 × 10 −31 kg )
32
(6.626 × 10 −34 J ⋅ s )
3
( 4.00 eV ) (1.60 × 10 −19 J eV )
g ( E ) = 8.50 × 10 46 m −3 ⋅ J −1 = 1.36 × 10 28 m −3 ⋅ eV−1
So, the total number of electrons is
N = ⎡⎣g ( E )⎤⎦ ( ∆E ) V = (1.36 × 10 28 m −3 ⋅ eV−1 ) ( 0.025 0 eV) (1.00 × 10 −9 m 3 )
= 3.40 × 1017 electrons
*P43.37 (a)
(b)
The density of states at energy E is
g ( E ) = CE 1 2
Hence, the required ratio is
g ( 8.50 eV)
C ( 8.50 )
12
=
g ( 7.00 eV) C ( 7.00 )1 2
From Equation 43.22, the number of occupied states having energy E is
N (E) =
Hence, the required ratio is
= 1.10
CE 1 2
e
( E − EF )
k BT
+1
(8.50) ⎡ e(7.00−77.00) k T + 1⎤
⎢
⎥
N ( 7.00 eV) ( 7.00 )1 2 ⎣ e(8.50−7.00) k T + 1⎦
N ( 8.50 eV)
12
=
B
B
At T = 300 K, kBT = 4.14 × 10 −21 J = 0.025 9 eV,
N ( 8.50 eV)
And
(8.50) ⎡ 2.00 ⎤
⎢
⎥
N ( 7.00 eV) ( 7.00 )1 2 ⎣ e(1.50) 0.025 9 + 1⎦
N ( 8.50 eV)
= 1.47 × 10 −25
N ( 7.00 eV)
12
=
This result is vastly smaller than that in part (a). We conclude that very few states well
above the Fermi energy are occupied at room temperature.
532
P43.38
Chapter 43
Consider first the wave function in x. At x = 0 and x = L , ψ = 0.
Therefore,
sin kx L = 0
and
kx L = π , 2 π , 3π , . . .
Similarly,
sin ky L = 0
and
ky L = π , 2 π , 3π , . . .
sin kz L = 0
and
kz L = π , 2 π , 3π , . . .
⎛n π x⎞ ⎛n π y⎞ ⎛n π z⎞
ψ = A sin ⎜ x ⎟ sin ⎜ y ⎟ sin ⎜ z ⎟
⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠
From
∂ 2 ψ ∂ 2 ψ ∂ 2 ψ 2 me
+
+
= 2 (U − E ) ψ,
∂ x 2 ∂ y2 ∂ z 2
⎛ nx2π 2 ny2π 2 nz2π 2 ⎞
2 me
⎜ − L2 − L2 − L2 ⎟ ψ = 2 ( − E ) ψ
⎝
⎠
we have inside the box, where U = 0,
E=
2 π 2 2
(nx + ny2 + nz2 ) nx , ny , nz = 1, 2, 3, . . .
2 me L2
Outside the box we require ψ = 0.
The minimum energy state inside the box is
Section 43.6
P43.39
nx = ny = nz = 1, with E =
Electrical Conduction in Metals, Insulators, and Semiconductors
Eg = 1.14 eV for Si
(a)
hf = 1.14 eV = (1.14 eV ) (1.60 × 10 −19 J eV ) = 1.82 × 10 −19 J
c = λ f; λ =
(b)
P43.40
3 2π 2
2 me L2
so
f ≥ 2.75 × 1014 Hz
c 3.00 × 108 m s
=
= 1.09 × 10 −6 m = 1.09 µ m (in the infrared region)
f
2.75 × 1014 Hz
Photons of energy greater than 2.42 eV will be absorbed. This means wavelength shorter than
−34
8
hc ( 6.626 × 10 J ⋅ s) ( 3.00 × 10 m s)
= 514 nm
=
E
2.42 × 1.60 × 10 −19 J
All the hydrogen Balmer lines except for the red line at 656 nm will be absorbed.
λ=
*P43.41 If λ ≤ 1.00 × 10 −6 m, then photons of sunlight have energy
E≥
−34
8
⎞
1 eV
hc ( 6.626 × 10 J ⋅ s) ( 3.00 × 10 m s) ⎛
=
⎜
⎟ = 1.24 eV
−6
−19
⎝ 1.60 × 10 J ⎠
1.00 × 10 m
λmax
Thus, the energy gap for the collector material should be Eg ≤ 1.24 eV . Since Si has
an energy gap Eg ≈ 1.14 eV, it can absorb nearly all of the photons in sunlight. Therefore,
Si is an appropriate material for a solar collector.
P43.42
Eg =
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
J ≈ 1.91 eV
λ
650 × 10 −9 m
Molecules and Solids
P43.43
533
If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here,
(hf )max =
hc
= 5.5 eV:
λmin
λmin =
(6.626 × 10 −34 J ⋅ s ) (3.00 × 108 m s )
hc
=
5.5 eV
( 5.5 eV ) (1.60 × 10 −19 J eV )
λmin = 2.26 × 10 −7 m = 226 nm
P43.44
In the Bohr model we replace ke by
ke
and me by m *. Then the radius of the first Bohr orbit,
κ
a0 =
2
in hydrogen, changes to
me k e e 2
a′ =
⎛ me ⎞
⎛m ⎞
⎛ me ⎞
2κ
2
= ⎜ e ⎟κ a0 = ⎜
κ
=
⎟11.7 ( 0.052 9 nm ) = 2.81 nm
⎜
⎟
2
2
m * k e e ⎝ m * ⎠ me k e e ⎝ m * ⎠
⎝ 0.220 me ⎠
The energy levels are in hydrogen En = −
En′ = −
⎛ m * ⎞ En
ke e 2 1
ke e 2
= −⎜
=
−
2
κ 2a ′ n
κ 2 ( m e m * ) κ a0
⎝ me ⎟⎠ κ 2
For n = 1, E1′ = −0.220
Section 43.7
P43.45
ke e 2 1
and here
2a0 n 2
13.6 eV
= −0.021 9 eV .
11.72
Semiconductor Devices
(
I = I0 e (
e ∆V ) kBT
)
−1
Thus,
and
∆V =
At T = 300 K,
∆V =
ee( ∆V ) kBT = 1 +
k BT ⎛
I ⎞
ln ⎜1 + ⎟
e
⎝ I0 ⎠
(1.38 × 10
−23
I
I0
J K ) ( 300 K )
1.60 × 10
−19
C
⎛
⎛
I ⎞
I ⎞
ln ⎜1 + ⎟ = ( 25.9 mV) ln ⎜1 + ⎟
⎝ I0 ⎠
⎝ I0 ⎠
(a)
If I = 9.00 I 0 ,
∆V = ( 25.9 mV) ln (10.0 ) = 59.5 mV
(b)
If I = −0.900 I 0 ,
∆V = ( 25.9 mV) ln ( 0.100 ) = −59.5 mV
The basic idea behind a semiconductor device is that a large current or charge can be
controlled by a small control voltage.
P43.46
The voltage across the diode is about 0.6 V. The voltage drop across the resistor is
( 0.025 A ) (150 Ω ) = 3.75 V. Thus, ε − 0.6 V − 3.8 V = 0 and ε = 4.4 V .
P43.47
First, we evaluate I 0 in I = I 0 ee( ∆V ) kBT − 1 , given that I = 200 mA when ∆V = 100 mV and
T = 300 K.
(
)
−19
e ( ∆V ) (1.60 × 10 C ) ( 0.100 V )
200 mA
I
= 3.86
= 4.28 mA
=
= 3.86 so I 0 = e( ∆V ) kBT
−23
k BT
−1 e −1
e
(1.38 × 10 J K ) (300 K )
If ∆V = −100 mV,
(
I = I0 e (
e ∆V ) kBT
)
e ( ∆V )
= −3.86; and the current will be
k BT
− 1 = ( 4.28 mA) ( e−3.86 − 1) = −4.19 mA
P43.48
Chapter 43
(a)
The currents to be plotted are
I D = (10
−6
A)(e
∆V 0.025 V
− 1) ,
2.42 V − ∆V
IW =
745 Ω
The two graphs intersect at
∆V = 0.200 V. The currents
are then
I D = (10
−6
A)(e
0.200 V 0.025 V
− 1)
Diode and Wire Currents
Current (mA)
534
20
Diode
Wire
10
0
0
0.1
= 2.98 mA
∆V(volts)
0.2
FIG. P43.48
2.42 V − 0.200 V
IW =
= 2.98 mA. They agree to three digits.
745 Ω
∴ I D = IW = 2.98 mA
(b)
∆V
0.200 V
=
= 67.1 Ω
ID
2.98 × 10 −3 A
(c)
d ( ∆V ) ⎡ dI D ⎤
⎡ 10 −6 A 0.200 V 0.025 V ⎤
=⎢
=
⎥
⎥ = 8.39 Ω
⎢ 0.025 V e
dI D
⎦
⎣
⎣ d ( ∆V ) ⎦
−1
Section 43.8
P43.49
−1
Superconductivity
By Faraday’s law (from Chapter 32)
∆Φ B
∆I
∆B
=L
=A
∆t
∆t
∆t
A ( ∆B)
π ( 0.010 0 m ) ( 0.020 0 T)
∆I =
=
= 203 A
L
3.10 × 10 −8 H
The direction of the induced current is such as to maintain the B–field through the ring.
Thus,
*P43.50 (a)
(b)
2
In the definition of resistance ∆V = IR, if R is
zero then ∆V = 0 for any value of current.
The graph shows a direct proportionality with
resistance given by the reciprocal of the slope:
1 ∆I
(155 − 57.8 ) mA
Slope = =
=
= 43.1 Ω −1
R ∆V ( 3.61 − 1.356 ) mV
R = 0.023 2 Ω
(c)
Expulsion of magnetic flux and therefore fewer
current-carrying paths could explain the
decrease in current.
FIG. P43.50
0.3
Molecules and Solids
P43.51
(a)
(b)
535
See the figure at right.
For a surface current around the outside of the cylinder as shown,
−2
B ( 0.540 T) ( 2.50 × 10 m )
N µ0 I
= 10.7 kA
=
B=
or NI =
µ0
( 4 π × 10−7 ) T ⋅ m A
FIG. P43.51
Additional Problems
P43.52
For the N 2 molecule, k = 2 297 N m , m = 2.32 × 10 −26 kg, r = 1.20 × 10 −10 m, µ =
m
2
2
k
= 4.45 × 1014 rad s , I = µ r 2 = (1.16 × 10 −26 kg ) (1.20 × 10 −10 m ) = 1.67 × 10 −46 kg ⋅ m 2
µ
For a rotational state sufficient to allow a transition to the first exited vibrational state,
−46
14
2
2 Iω 2 (1.67 × 10 ) ( 4.45 × 10 )
=
= 1 410.
J ( J + 1) = ω so J ( J + 1) =
2I
1.055 × 10 −34
ω=
Thus J = 37 .
*P43.53 (a)
(b)
Since the interatomic potential is the same for both molecules, the spring constant is the
same.
(12 u ) (16 u )
(14 u ) (16 u )
1 k
where µ12 =
Then f =
= 6.86 u and µ14 =
= 7.47 u.
12 u + 16 u
14 u + 16 u
2π µ
Therefore,
1
1
µ
k
k ⎛ µ12 ⎞
6.86 u
13
=
f14 =
= 6.15 × 1013 Hz
⎜
⎟ = f12 12 = ( 6.42 × 10 Hz )
2 π µ14 2 π µ12 ⎝ µ14 ⎠
µ14
7.47 u
The equilibrium distance is the same for both molecules.
⎛µ ⎞
⎛µ ⎞
I14 = µ14 r 2 = ⎜ 14 ⎟ µ12 r 2 = ⎜ 14 ⎟ I12
⎝ µ12 ⎠
⎝ µ12 ⎠
7.47 u ⎞
1.46 × 10 −46 kg ⋅ m 2 ) = 1.59 × 10 −46 kg ⋅ m 2
I14 = ⎛
⎝ 6.86 u ⎠ (
(c)
The molecule can move to the ( v = 1, J = 9 ) state or to the ( v = 1, J = 11) state. The
energy it can absorb is either
2 ⎤
2 ⎤ ⎡⎛ 1 ⎞
hc ⎡⎛ 1 ⎞
∆E =
= ⎢⎜1 + ⎟ hf14 + 9 ( 9 + 1)
⎥ − ⎢⎜ 0 − ⎟ hf14 + 10 (10 + 1)
⎥
λ ⎣⎝ 2 ⎠
2 I14 ⎦
2 I14 ⎦ ⎣⎝ 2 ⎠
or
∆E =
2 ⎤
2 ⎤ ⎡⎛ 1 ⎞
hc ⎡⎛ 1 ⎞
= ⎢⎜1 + ⎟ hf14 + 11(11 + 1)
⎥ − ⎢⎜ 0 + ⎟ hf14 + 10 (10 + 1)
⎥
λ ⎣⎝ 2 ⎠
2 I14 ⎦ ⎣⎝ 2 ⎠
2 I14 ⎦
The wavelengths it can absorb are then
c
c
λ=
or λ =
f14 + 11 ( 2 π I14 )
f14 − 10 ( 2π I14 )
These are:
λ=
and λ =
3.00 × 108 m s
= 4.96 µ m
6.15 × 1013 Hz − ⎡⎣10 (1.055 × 10 −34 J ⋅ s ) ⎤⎦ ⎡⎣ 2π (1.59 × 10 −46 kg ⋅ m 2 ) ⎤⎦
3.00 × 108 m s
= 4.79 µ m
6.15 × 1013 Hz + ⎡⎣11 (1.055 × 10 −34 J ⋅ s ) ⎤⎦ ⎡⎣ 2π (1.59 × 10 −46 kg ⋅ m 2 ) ⎤⎦
536
P43.54
P43.55
Chapter 43
With 4 van der Waals bonds per atom pair or 2 electrons per atom, the total energy of the solid is
⎛ 6.02 × 10 23 atoms ⎞
E = 2 (1.74 × 10 −23 J atom ) ⎜
⎟ = 5.23 J g
4.00 g
⎝
⎠
⎛
1⎞
∆Emax = 4.5 eV = ⎜ v + ⎟ ω
⎝
2⎠
8.25 > 7.5
P43.56
P43.57
( 4.5 eV ) (1.6 × 10 −19 J eV )
so
(1.055 × 10
v=7
−34
1
≥ ⎛v + ⎞
⎝
2⎠
J ⋅ s ) (8.28 × 10 s )
14
−1
3
1
hf + 4.48 eV = hf + 3.96 eV is the
2
2
depth of the well below the dissociation point. We see hf = 0.520 eV, so the depth of the well is
1
1
hf + 4.48 eV = ( 0.520 eV ) + 4.48 eV = 4.74 eV .
2
2
Suppose it is a harmonic-oscillator potential well. Then,
ke e 2 B
+ m.
r
r
The total potential energy has its minimum value U 0 at the equilibrium spacing, r = r0 . At this
dU
point,
= 0,
dr r =r0
The total potential energy is given by Equation 43.17: U total = −α
or
dU
dr
=
r =r0
d ⎛ ke e 2 B ⎞
+ m⎟
⎜−α
dr ⎝
r
r ⎠
=α
r =r0
k e2
Thus, B = α e r0m−1
m
Substituting this value of B into U total ,
*P43.58 (a)
ke e2 mB
− m+1 = 0
r02
r0
U 0 = −α
⎛1⎞
ke e 2
k e2
k e2 ⎛ 1 ⎞
+ α e r0m−1 ⎜ m ⎟ = −α e ⎜1 − ⎟
r0
m
r0 ⎝ m ⎠
⎝ r0 ⎠
ke e 2 B
+ m has its minimum value at
r
r
dU
the equilibrium spacing, r = r0 . At this point, F = −
= 0, or
dr r =r0
The total potential energy −α
F =−
d ⎛ ke e 2 B ⎞
k e2 mB
+ m ⎟ = −α e 2 + m+1 = 0
⎜−α
dr ⎝
r
r ⎠ r =r
r0
r0
0
k e2
Thus, B = α e r0m −1 .
m
m −1
⎤
k e2
m
k e2
k e2 ⎡ ⎛ r
Substituting this value of B into F, F = −α e 2 + m +1 α e r0m −1 = −α e 2 ⎢1 − 0 ⎞ ⎥ .
r
r
m
r ⎣ ⎝ r⎠ ⎦
(b)
Let r = r0 + x so
F = −α
≈ −α
ke e
r2
2
r0 = r – x
⎡ ⎛ r − x⎞
⎢1 − ⎝
r ⎠
⎣
m −1
Then assuming x is small we have
⎤
ke e 2
=
−
α
⎥
r2
⎦
m −1
⎡ ⎛
x⎞ ⎤
ke e 2
−
−
−
1
1
≈
α
⎥
⎢ ⎝
r⎠ ⎦
r2
⎣
⎡1 − 1 + ( m − 1) x ⎤
⎢⎣
r ⎥⎦
ke e 2
( m − 1) x
r03
This is of the form of Hooke’s law with spring constant K = kea e2(m – 1)/r03.
(c)
Section 38.5 on electron diffraction gives the interatomic spacing in NaCl as (0.562 737 nm)/2.
Other problems in this chapter give the same information, or we could calculate it from the
statement in the chapter text that the ionic cohesive energy for this crystal is –7.84 eV.
The stiffness constant is then
ke e 2
8.99 × 10 9 N ⋅ m 2 (1.6 × 10 −19 C)2 (8 − 1)
(
−
1
)
=
1
.
7476
= 127 N/m
m
r03
C2 (2.81 × 10 −10 m)3
The vibration frequency of a sodium ion within the crystal is
1 K
1
127 N/m
f =
=
= 9.18 THz
2π m 2π 23.0 × 1.66 × 10 −27 kg
K =α
Molecules and Solids
P43.59
(a)
d
dU
Ax −3 − Bx −1 ) = −3 Ax −4 + Bx −2 = 0
=0:
(
dx
dx
x → ∞ describes one equilibrium position, but the stable equilibrium position is at
3 Ax0−2 = B.
For equilibrium,
3 ( 0.150 eV ⋅ nm 3 )
3A
=
B
x0 =
(b)
537
3.68 eV ⋅ nm
The depth of the well is given by
= 0.350 nm
U0 = U
x = x0
=
A B
AB 3 2
BB1 2
− = 32 32 − 12 12
3
x0 x0 3 A
3 A
2 ( 3.68 eV ⋅ nm )
2 B3 2
=−
= −7.02 eV
12
32
33 2 A1 2
3 ( 0.150 eV ⋅ nm 3 )
32
U 0 = U x= x = −
0
(c)
Fx = −
dU
= 3 Ax −4 − Bx −2
dx
To find the maximum force, we determine finite xm such that
dF
dx
= 0.
x = xm
⎛ 6A ⎞
= 0 so that xm = ⎜ ⎟
⎝ B⎠
12
−5
−3
Thus,
⎡⎣ −12 Ax + 2 Bx ⎤⎦ x = x
m
Then
B
B
B2
( 3.68 eV ⋅ nm )
Fmax = 3 A ⎛ ⎞ − B ⎛ ⎞ = −
=−
⎝ 6A ⎠
⎝ 6A ⎠
12 A
12 ( 0.150 eV ⋅ nm 3 )
or
⎛ 1.60 × 10 −19
Fmax = −7.52 eV nm ⎜
1 eV
⎝
2
P43.60
(a)
For equilibrium,
2
J ⎞ ⎛ 1 nm ⎞
−9
⎟ ⎜ −9 ⎟ = −1.20 × 10 N = −1.20 nN
⎝
⎠
10
m
⎠
d
( Ax −3 − Bx −1 ) = −3Ax −4 + Bx −2 = 0
dx
dU
= 0:
dx
x → ∞ describes one equilibrium position, but the stable equilibrium position is at
3 Ax0−2 = B
(b)
(c)
or
x0 =
3A
B
The depth of the well is given by U 0 = U
x = x0
=
A B
AB 3 2
BB1 2
B3
− = 3 2 3 2 − 1 2 1 2 = −2
.
3
3 A
27A
x0 x0 3 A
dU
= 3Ax −4 − Bx −2
dx
To find the maximum force, we determine finite xm such that
Fx = −
dFx
dx
⎛ B⎞
⎛ B⎞
B2
= 0 then Fmax = 3A ⎜ ⎟ − B ⎜ ⎟ = −
⎝ 6A ⎠
⎝ 6A ⎠
12 A
2
−5
x = xm
−3
= ⎡⎣ −12 Ax + 2 Bx ⎤⎦ x = x
0
538
P43.61
Chapter 43
(a)
At equilibrium separation, r = re ,
dU
dr
r =re
− a r −r
− a r −r
= −2 aB ⎡⎣e ( e 0 ) − 1⎤⎦ e ( e 0 ) = 0
We have neutral equilibrium as re → ∞ and stable equilibrium at e− a( re − r0 ) = 1,
or
re = r0
(b)
At r = r0 , U = 0. As r → ∞, U → B. The depth of the well is B .
(c)
We expand the potential in a Taylor series about the equilibrium point:
U ( r ) ≈ U ( r0 ) +
U (r ) ≈ 0 + 0 +
(d)
dU
dr
r =r0
(r − r0 ) +
1 d 2U
2 dr 2
(r − r0 )
2
r =r0
1
2
2
(−2 Ba) ⎡⎣−ae−2 ( r−r0 ) − ae−( r−r0 ) (e−2 ( r−r0 ) − 1)⎤⎦r−r (r − r0 ) ≈ Ba 2 (r − r0 )
2
0
This is of the form
1 2 1
2
kx = k ( r − r0 )
2
2
for a simple harmonic oscillator with
k = 2 Ba 2
Then the molecule vibrates with frequency
f=
The zero-point energy is
1
1
ha B
ω = hf =
2
2
π 8µ
1
2π
k
a
=
µ 2π
2B
a
=
µ
π
Therefore, to dissociate the molecule in its ground state requires energy B −
B
2µ
ha B
.
π 8µ
Molecules and Solids
P43.62
T =0
T = 0.1TF
T = 0.2TF
539
T = 0.5TF
E
EF
e⎣
0
e−∞
1.00
e−10.0
1.000
e−5.00
0.993
e−2.00
0.881
0.500 e−∞
1.00
e−5.00
0.993
e−2.50
0.924
e−1.00
0.731
0.600 e−∞
1.00
e−4.00
0.982
e−2.00
0.881
e−0.800
0.690
0.700 e−∞
1.00
e−3.00
0.953
e−1.50
0.818
e−0.600
0.646
0.800 e−∞
1.00
e−2.00
0.881
e−1.00
0.731
e−0.400
0.599
0.900 e−∞
1.00
e−1.00
0.731
e−0.500
0.622
e−0.200
0.550
1.00
e0
0.500
e0
0.500
e0
0.500
e0
0.500
1.10
e+∞
0.00
e1.00
0.269
e0.500
0.378
e0.200
0.450
1.20
e+∞
0.00
e2.00
0.119
e1.00
0.269
e0.400
0.401
1.30
e+∞
0.00
e3.00
0.047 4
e1.50
0.182
e0.600
0.354
1.40
e+∞
0.00
e4.00
0.018 0
e2.00
0.119
e0.800
0.310
1.50
e+∞
0.00
e5.00
0.006 69 e2.50
⎡( E EF )−1⎤⎦(TF T )
f (E)
⎡( E EF )−1⎤⎦(TF T )
e⎣
f (E)
⎡( E EF )−1⎤⎦(TF T )
e⎣
f (E)
⎡( E EF )−1⎤⎦(TF T )
e⎣
0.075 9 e1.00
FIG. P43.62
ANSWERS TO EVEN PROBLEMS
P43.2
4.3 eV
P43.4
(a) 1.28 eV
P43.6
(a) 0.0148 eV
P43.8
12.2 × 10–27 kg; 12.4 × 10–27 kg; They agree, because the small apparent difference can be
attributed to uncertainty in the data.
(b) σ = 0.272 nm, ∈= 4.65 eV
(c) 6.55 nN
(d) 576 N m
(b) 83.8 mm
f (E)
0.269
540
Chapter 43
P43.10
1.46 × 10 −46 kg ⋅ m 2
P43.12
(a) 1.81 × 10 −45 kg ⋅ m 2
P43.14
(a) 12.1 pm
P43.16
(a) 0, 364 µ eV, 1.09 meV
(b) 98.2 meV, 295 meV, 491 meV
P43.18
(a) 472 µ m
(c) 0.715 µ m
P43.20
(a) 4.60 × 10–48 kg · m2 (b) 1.32 ×1014 Hz
P43.22
6.25 × 109
P43.24
(a) ~1017
P43.26
(a) 0.444 nm, 0.628 nm, 0.769 nm
P43.28
(a) The Fermi energy is proportional to the spatial concentration of free electrons to the
two-thirds power. (c) 6.06; copper by 3.33 times; it agrees with the equation because
6.062/3 = 3.33.
P43.30
0.938
P43.32
(a) 2.54 × 1028 electron/m3 (b) 3.15 eV
P43.34
2%
P43.36
3.40 × 1017 electrons
P43.38
See the solution.
P43.40
All of the Balmer lines are absorbed, except for the red line at 656 nm, which is transmitted.
P43.42
1.91 eV
P43.44
−0.021 9 eV, 2.81 nm
P43.46
4.4 V
P43.48
(a) At ∆V = 0.200 V, ID = IW = 2.98 mA, agreeing to three digits.
P43.50
(a) In the definition of resistance ∆V = IR, if R is zero then ∆V = 0 for any value of current.
(b) The graph shows direct proportionality with resistance 0.023 2 Ω. (c) Expulsion of magnetic
flux and therefore fewer current-carrying paths could explain the decrease in current.
P43.52
37
P43.54
5.23 J g
P43.56
4.74 eV
P43.58
(c) 9.18 THz
P43.60
(a) x0 =
P43.62
See the solution.
(b) 1.62 cm
(b) 9.23 pm
(c) HI is more loosely bound since it has the smaller k value.
(b) 473 µ m
(c) 0.074 1 nm
(b) ~10 5 m 3
3A
B
(b) −2
B3
27 A
(c) −
B2
12 A
(b) 67.1 Ω
(c) 8.39 Ω
44
Nuclear Structure
CHAPTER OUTLINE
44.1
44.2
44.3
44.4
44.5
44.6
44.7
44.8
Some Properties of Nuclei
Nuclear Binding Energy
Nuclear Models
Radioactivity
The Decay Processes
Natural Radioactivity
Nuclear Reactions
Nuclear Magnetic Resonance
and Magnetic Resonance
Imagining
ANSWERS TO QUESTIONS
Q44.1
Because of electrostatic repulsion between the
positively-charged nucleus and the +2e alpha particle.
To drive the a -particle into the nucleus would require
extremely high kinetic energy.
*Q44.2 (a)
(b)
(c)
X has a mass number less by 2 than the others.
The ranking is W = Y = Z > X.
Y has a greater atomic number, because a neutron
in the parent nucleus has turned into a proton.
X has an atomic number less by two than W,
so the ranking is Y > W = Z > X.
Y has one fewer neutron compared to the parent
nucleus W, and X has two fewer neutrons than W.
The ranking is W = Z > Y > X.
Q44.3
The nuclear force favors the formation of neutron-proton pairs, so a stable nucleus cannot be too
far away from having equal numbers of protons and neutrons. This effect sets the upper boundary
of the zone of stability on the neutron-proton diagram. All of the protons repel one another
electrically, so a stable nucleus cannot have too many protons. This effect sets the lower boundary
of the zone of stability.
Q44.4
Nuclei with more nucleons than bismuth-209 are unstable because the electrical repulsion forces
among all of the protons is stronger than the nuclear attractive force between nucleons.
Q44.5
Nucleus Y will be more unstable. The nucleus with the higher binding energy requires more
energy to be disassembled into its constituent parts.
Q44.6
Extra neutrons are required to overcome the increasing electrostatic repulsion of the protons.
The neutrons participate in the net attractive effect of the nuclear force, but feel no Coulomb
repulsion.
*Q44.7 (i) Answer (a). The liquid drop model gives a simpler account of a nuclear fission reaction,
including the energy released and the probable fission product nuclei.
(ii) Answer (b). The shell model predicts magnetic moments by necessarily describing the spin
and orbital angular momentum states of the nucleons.
(iii) Answer (b). Again, the shell model wins when it comes to predicting the spectrum of an
excited nucleus, as the quantum model allows only quantized energy states, and thus only specific
transitions.
541
542
Q44.8
Chapter 44
The statement is false. Both patterns show monotonic decrease over time, but with very different
shapes. For radioactive decay, maximum activity occurs at time zero. Cohorts of people now
living will be dying most rapidly perhaps forty years from now. Everyone now living will be
dead within less than two centuries, while the mathematical model of radioactive decay tails off
exponentially forever. A radioactive nucleus never gets old. It has constant probability of decay
however long it has existed.
*Q44.9 (i) Answer (b). Since the samples are of the same radioactive isotope, their half-lives are the same.
(ii) Answer (b). When prepared, sample G has twice the activity (number of radioactive
decays per second) of sample H. After 5 half-lives, the activity of sample G is decreased
by a factor of 25, and after 5 half-lives the activity of sample H is decreased by a factor of 25.
So after 5 half-lives, the ratio of activities is still 2:1.
Q44.10 After one half-life, one half the radioactive atoms have decayed. After the second half-life, one
1 1 3
half of the remaining atoms have decayed. Therefore + = of the original radioactive atoms
2 4 4
have decayed after two half-lives.
Q44.11
Long-lived progenitors at the top of each of the three natural radioactive series are the sources
of our radium. As an example, thorium-232 with a half-life of 14 Gyr produces radium-228 and
radium-224 at stages in its series of decays, shown in Figure 44.17.
*Q44.12 Answer (d). A free neutron decays into a proton plus an electron and an antineutrino. This
implies that a proton is more stable than a neutron, and in particular the proton has lower mass.
Therefore a proton cannot decay into a neutron plus a positron and a neutrino. This reaction
satisfies every conservation law except for energy conservation.
*Q44.13 The alpha particle and the daughter nucleus carry equal amounts of momentum in opposite
p2
directions. Since kinetic energy can be written as
, the small-mass alpha particle has much
2m
more of the decay energy than the recoiling nucleus.
Q44.14 Yes. The daughter nucleus can be left in its ground state or sometimes in one of a set of excited
states. If the energy carried by the alpha particle is mysteriously low, the daughter nucleus can
quickly emit the missing energy in a gamma ray.
*Q44.15 Answer (d). The reaction energy is the amount of energy released as a result of a nuclear reaction.
Equation 44.28 in the text implies that the reaction energy is (initial mass − final mass) c2.
The Q-value is taken as positive for an exothermic reaction.
*Q44.16 The samples would have started with more carbon-14 than we first thought. We would increase
our estimates of their ages.
5
5 3 1 1 3
5
Q44.17 Iz may have 6 values for I = , namely , , , − , − , and − . Seven Iz values are possible
2
2 2 2 2 2
2
for I = 3.
Nuclear Structure
543
*Q44.18 Answer (b). The frequency increases linearly with the magnetic field strength.
Q44.19 The decay of a radioactive nucleus at one particular moment instead of at another instant cannot
be predicted and has no cause. Natural events are not just like a perfect clockworks. In history,
the idea of a determinate mechanical Universe arose temporarily from an unwarranted wild
extrapolation of Isaac Newton’s account of planetary motion. Before Newton’s time [really you
can blame Pierre Simon de Laplace] and again now, no one thought of natural events as just like
a perfect row of falling dominos. We can and do use the word “cause” more loosely to describe
antecedent enabling events. One gear turning another is intelligible. So is the process of a hot
dog getting toasted over a campfire, even though random molecular motion is at the essence of
that process. In summary, we say that the future is not determinate. All natural events have
causes in the ordinary sense of the word, but not necessarily in the contrived sense of a cause
operating infallibly and predictably in a way that can be calculated. We have better reason now
than ever before to think of the Universe as intelligible. First describing natural events, and
second determining their causes form the basis of science, including physics but also scientific
medicine and scientific bread-baking. The evidence alone of the past hundred years of discoveries
in physics, finding causes of natural events from the photoelectric effect to x-rays and jets
emitted by black holes, suggests that human intelligence is a good tool for figuring out how
things go. Even without organized science, humans have always been searching for the causes of
natural events, with explanations ranging from “the will of the gods” to Schrödinger’s equation.
We depend on the principle that things are intelligible as we make significant strides towards
understanding the Universe. To hope that our search is not futile is the best part of human nature.
SOLUTIONS TO PROBLEMS
Section 44.1
P44.1
Some Properties of Nuclei
An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water
molecule there are eight neutrons and ten protons.
So protons and neutrons are nearly equally numerous in your body, each contributing mass
(say) 35 kg:
⎛ 1 nucleon ⎞
35 kg ⎜
~10 28 protons
⎝ 1.67 × 10 −27 kg ⎟⎠
and ~10 28 neutrons
The electron number is precisely equal to the proton number, ~10 28 electrons .
544
P44.2
Chapter 44
1
m v 2 = q∆V
2
mv 2
= qv B
r
and
2 m ∆V
=
qB 2
r=
2 m ∆V = qr 2 B 2 :
2 (1 000 V )
m
(1.60 × 10 −19 C) ( 0.200 T )2
(
r = 5.59 × 1011 m
(a)
For 12 C, m = 12 u
kg
)
m
(
r = 5.59 × 1011 m
and
kg
)
12 (1.66 × 10 −27 kg )
r = 0.078 9 m = 7.89 cm
(
r = 5.59 × 1011 m
For 13C:
kg
)
13 (1.66 × 10 −27 kg )
r = 0.082 1 m = 8.21 cm
(b)
With
the ratio gives
r1 =
r1
=
r2
2 m1 ∆V
qB 2
and
r2 =
2 m2 ∆V
qB 2
m1
m2
r1 7.89 cm
=
= 0.961
r2 8.21 cm
m1
12 u
=
= 0.961
m2
13 u
and
so they do agree.
*P44.3
(a)
Let V represent the volume of the tank. The number of moles present is
n=
PV 1.013 × 10 5 N V mol ⋅ K
=
= 44.6 V mol/m 3
RT 8.314 N ⋅ m m 2 273 K
Then the number of molecules is 6.023 × 1023 × 44.6 V /m3.
3
4
8π ⎛ 10 −10 m ⎞
= 1.047 × 10 −30 m 3 .
The volume of one molecule is 2 π r 3 =
2 ⎟⎠
3
3 ⎜⎝
The volume of all the molecules is 2.69 × 1025 V (1.047 × 10−30) = 2.81 × 10 −5 V.
So the fraction of the volume occupied by the hydrogen molecules is 2.81 × 10−5 .
An atom is precisely one half of a molecule.
(b)
nuclear volume
=
atomic volume
3
⎛ 1.20 × 10 −15 m ⎞
π r3
−14
=
⎜⎝ 0.5 × 10 −10 m ⎟⎠ = 1.38 × 10
4 π ( d / 2 )3
3
4
3
In linear dimension, the nucleus is small inside the atom in the way a fat strawberry is small
inside the width of the Grand Canyon. In terms of volume, the nucleus is really small.
Nuclear Structure
P44.4
545
Eα = 7.70 MeV
−19
9
4 k Ze2 2 k Ze2 2 (8.99 × 10 ) ( 79 ) (1.60 × 10 )
= 29.5 × 10 −15 m = 29.5 fm
= e 2 = e
=
mv
Eα
7.70 (1.60 × 10 −13 )
2
(a)
dmin
(b)
The de Broglie wavelength of the α is
λ=
P44.5
h
=
mα vα
h
=
2 mα Eα
6.626 × 10 −34
2 ( 6.64 × 10
−27
) 7.70 (1.60 × 10 )
−13
= 5.18 × 10 −15 m = 5.18 fm
(c)
Since λ is much less than the distance of closest approach , the α may be considered a
particle.
(a)
The initial kinetic energy of the alpha particle must equal the electrostatic potential energy
at the distance of closest approach.
k qQ
K i = Uf = e
rmin
9
2
2
−19
ke qQ (8.99 × 10 N ⋅ m C ) ( 2 ) ( 79 ) (1.60 × 10 C )
=
= 4.55 × 10 −13 m
Ki
( 0.500 MeV ) (1.60 × 10 −13 J MeV )
2
rmin =
(b)
Since K i =
vi =
P44.6
P44.7
2 ke qQ
=
mα rmin
2 (8.99 × 10 9 N ⋅ m 2 C2 ) ( 2 ) ( 79 ) (1.60 × 10 −19 C )
( 4.00 u ) (1.66 × 10 −27 kg u ) ( 3.00 × 10 −13 m )
(a)
r = r0 A1 3 = (1.20 × 10 −15 m ) ( 4 )
(b)
r = r0 A1 3 = (1.20 × 10 −15 m ) ( 238)
= 1.90 × 10 −15 m
13
13
= 7.44 × 10 −15 m
The number of nucleons in a star of two solar masses is
A=
Therefore
P44.8
1
k qQ
mα vi2 = e ,
2
rmin
2 (1.99 × 10 30 kg )
1.67 × 10 −27 kg nucleon
= 2.38 × 10 57 nucleons
r = r0 A1 3 = (1.20 × 10 −15 m ) ( 2.38 × 10 57 )
13
= 16.0 km
4
4
3
V = π r 4 = π ( 0.021 5 m ) = 4.16 × 10 −5 m 3
3
3
We take the nuclear density from Example 44.2
m = ρV = ( 2.3 × 1017 kg m 3 ) ( 4.16 × 10 −5 m 3 ) = 9.57 × 1012 kg
(9.57 × 10 kg )
m1 m2
= (6.67 × 10 −11 N ⋅ m 2 kg 2 )
r2
(1.00 m )2
12
and
F=G
F = 6.11 × 1015 N toward the other ball.
2
2
= 6.04 × 10 6 m s
546
Chapter 44
Section 44.2
P44.9
Nuclear Binding Energy
Using atomic masses as given in the table in the text,
(a)
For 21 H:
−2.014 102 + 1(1.008 665) + 1(1.007 825)
2
Eb
931.5 MeV ⎞
= 1.11 MeV nucleeon
= ( 0.001194 u ) ⎛
⎝
⎠
A
u
(b)
For 42 He:
(c)
For
56
26
Fe:
2 (1.008 665) + 2 (1.007 825) − 4.002 603
4
Eb
= 0.007 59 uc 2 = 7.07 MeV nucleon
A
30 (1.008 665) + 26 (1.007 825) − 55.934 942 = 0.528 u
Eb 0.528
=
= 0.009 44 uc 2 = 8.79 MeV nucleon
A
56
(d)
For
238
92
U:
146 (1.008 665) + 92 (1.007 825) − 238.050 783 = 1.934 2 u
Eb 1.934 2
=
= 0.008 13 uc 2 = 7.57 MeV nucleon
238
A
∆M = ZmH + Nmn − M
*P44.10
Nuclei
Z
N
M in u
55
25
26
27
30
30
32
54.938 050
55.934 942
58.933 200
Mn
Fe
59
Co
56
∴ 56 Fe has a greater
P44.11
(a)
P44.12
0.517 5
0.528 46
0.555 35
8.765
8.790
8.768
Eb
than its neighbors. This tells us finer detail than is shown in Figure 44.5.
A
A−Z
is greatest for 139
55 Cs and is equal to 1.53.
Z
139
La has the largest binding energy per nucleon of 8.378 MeV.
The neutron-to-proton ratio
(b)
(c)
Eb ∆M ( 931.5)
=
A
A
Eb
∆M in u
in MeV
A
139
Cs with a mass of 138.913 u. We locate the nuclei carefully on Figure 44.4, the
neutron−proton plot of stable nuclei. Cesium appears to be farther from the center of the
zone of stability. Its instability means extra energy and extra mass.
Use Equation 44.2.
Then for
and for
23
12
23
11
Na,
Mg,
Eb
= 8.11 MeV nucleon
A
Eb
= 7.90 MeV nucleon
A
The binding energy per nucleon is greater for 23
11 Na by 0.210 MeV . There is less proton
23
repulsion in Na . It is the more stable nucleus.
Nuclear Structure
P44.13
The binding energy of a nucleus is Eb ( MeV ) = ⎡⎣ ZM ( H ) + Nmn − M ( ZA X ) ⎤⎦ ( 931.494 MeV u )
For
15
8
O: Eb = ⎣⎡8 (1.007 825 u ) + 7 (1.008 665 u ) − 15.003 065 u ⎤⎦ ( 931.494 MeV u ) = 111.96 MeV
For
15
7
N: Eb = ⎣⎡7 (1.007 825 u ) + 8 (1.008 665 u ) − 15.000 109 u ⎤⎦ ( 931.494 MeV u ) = 115.49 MeV
Therefore, the binding energy of
P44.14
547
(a)
(b)
15
7
N is larger by 3.54 MeV .
R = r0 A1 3 = (1.20 × 10 −15 m ) ( 40 ) = 4.10 × 10 −15 m
The radius of the 40Ca nucleus is:
The energy required to overcome electrostatic repulsion is
2
9
2
2
−19
3keQ 2 3 (8.99 × 10 N ⋅ m C ) ⎡⎣ 20 (1.60 × 10 C ) ⎤⎦
U=
=
= 1.35 × 10 −11 J = 84.1 MeV
V
5R
5 ( 4.10 × 10 −15 m )
13
The binding energy of
40
20
Ca is
Eb = ⎡⎣ 20 (1.007 825 u ) + 20 (1.008 665 u ) − 39.962 591 u ⎤⎦ ( 931.5 MeV u ) = 342 MeV
(c)
P44.15
The nuclear force is so strong that the binding energy greatly exceeds the minimum energy
needed to overcome electrostatic repulsion.
43
Removal of a neutron from 20
Ca would result in the residual nucleus,
separation energy is Sn , the overall process can be described by
42
20
Ca. If the required
43
mass ( 20
Ca ) + Sn = mass ( 42
20 Ca ) + mass ( n )
Sn = ( 41.958 618 + 1.008 665 − 42.958 767 ) u = ( 0.008 516 u ) ( 931.5 MeV u ) = 7.93 MeV
Section 44.3
P44.16
(a)
(b)
P44.17
Nuclear Models
The first term overstates the importance of volume and the second term subtracts this
overstatement.
( 4 3) π R 3 = R .
R3
R
=
.
For spherical volume
For
cubical
volume
2
2
6R
6
4π R
3
The maximum binding energy or lowest state of energy is achieved by building “nearly”
spherical nuclei.
∆Eb = Ebf − Ebi
Eb
= 7.8 MeV
A
For
A = 200,
so
Ebi = 200 ( 7.8 MeV ) = 1 560 MeV
For
A ≈ 100,
so
Ebf = 2 (100 ) (8.7 MeV )
Eb
= 8.7 MeV
A
= 1 740 MeV
∆Eb = Ebf − Ebi :
Eb = 1 740 MeV − 1 560 MeV
≈ 200 MeV
FIG. P44.17
548
P44.18
Chapter 44
(a)
“Volume” term:
E1 = C1 A = (15.7 MeV ) ( 56 ) = 879 MeV
“Surface” term:
E2 = −C2 A 2 3 = − (17.8 MeV ) ( 56 )
“Coulomb” term:
E3 = −C3
“Asymmetry” term:
E4 = C 4
23
= −260 MeV
Z ( Z − 1)
( 26 ) ( 25)
= − ( 0.71 MeV )
= −121 MeV
13
A
( 56 )1 3
( A − 2 Z )2
A
= − ( 23.6 MeV )
( 56 − 52 )2
56
= −6.74 MeeV
Eb = 491 MeV
(b)
Section 44.4
P44.19
E
E1
E
E
= 179%; 2 = −53.0%; 3 = −24.6%; 4 = −1.37%
Eb
Eb
Eb
Eb
Radioactivity
dN
= −λ N
dt
so
λ=
1 ⎛ dN ⎞
−15
11
−4 −1
⎜−
⎟ = (1.00 × 10 ) (6.00 × 10 ) = 6.00 × 10 s
N ⎝ dt ⎠
T1 2 =
P44.20
P44.21
ln 2
= 1.16 × 10 3 s ( = 19.3 min )
λ
5
⎛ 1⎞
R = R0 e− λ t = (6.40 mCi ) e−(ln 2 8.04 d )(40.2 d ) = (6.40 mCi ) ( e− ln 2 ) = (6.40 mCi ) ⎜ 5 ⎟ = 0.200 mCi
⎝2 ⎠
(a)
From
R = R0 e− λ t ,
R
1
1 ⎞ ⎛ 10.0 ⎞
ln
λ = ln ⎛ 0 ⎞ = ⎛
= 5.58 × 10 −2 h −1 = 1.55 × 10 −5 s −1
t ⎝ R ⎠ ⎝ 4.00 h ⎠ ⎝ 8.00 ⎠
T1 2 =
ln 2
= 12.4 h
λ
R0 10.0 × 10 −3 Ci ⎛ 3.70 × 1010 s ⎞
=
= 2.39 × 1013 atoms
−5
⎜
⎟
⎝
⎠
λ
1.55 × 10 s
1 Ci
(b)
N0 =
(c)
R = R0 e− λ t = (10.0 mCi ) exp ( −5.58 × 10 −2 × 30.0 ) = 1.88 mCi
Nuclear Structure
*P44.22 From the law of radioactive decay N = N0 e−lt the decay rate is originally R0 = −dN/dt and
R = R0 e−lt
decreases according to R = −dN/dt = +N0 λ e−lt
Now λ =
ln 2
ln 2 1 ⎛ R0 ⎞
gives
= ln
T1/ 2
T1/ 2 t ⎝ R ⎠
T1/ 2 =
1 ⎛R ⎞
λ = ln ⎜ 0 ⎟
t ⎝ R⎠
ln(R0/R) = λ t
Then algebra to isolate the decay constant gives R0/R = elt
(ln 2)t
ln( R0 / R)
where t = ∆t is the time interval during which the activity decreases from R0 to R.
P44.23
The number of nuclei that decay during the interval will be N1 − N 2 = N 0 ( e− λ t1 − e− λ t2 ) .
ln 2 0.693
=
= 0.010 7 h −1 = 2.97 × 10 −6 s −1
T1 2 64.8 h
First we find λ:
λ=
and
N0 =
Substituting these values,
−( 0.010 7 h −1 )(10.0 h )
−( 0.010 7 h −1 )(12.0 h ) ⎤
N1 − N 2 = ( 4.98 × 1011 ) ⎡⎢ e
−e
⎣
⎦⎥
−1
4
R0 ( 40.0 µ Ci ) ( 3.70 × 10 s µ Ci )
= 4.98 × 1011 nuclei
=
λ
2.97 × 10 −6 s −1
Hence, the number of nuclei decaying during the interval is N1 − N 2 = 9.47 × 10 9 nuclei .
P44.24
The number of nuclei that decay during the interval will be N1 − N 2 = N 0 ( e− λ t1 − e− λ t2 ) .
First we find λ:
λ=
ln 2 − t T
−t T
e− λ t = e ( 1 2 ) = 2 1 2
R0T1 2
R
N0 = 0 =
λ
ln 2
so
and
Substituting in these values
N1 − N 2 =
*P44.25 The number remaining after time
is
ln 2
T1 2
R0T1 2
ln 2
(e
− λ t1
)
− e− λ t2 =
R0T1 2
ln 2
(2
− t1 T1 2
−2
− t 2 T1 2
)
T1/ 2 ln 2
=
2
2λ
N = N 0 e− λt = N 0 e− λ ln 2 / 2 λ = N 0 ( e− ln 2 )
1/ 2
⎛ 1⎞
= N0 ⎜ ⎟
⎝ 2⎠
1/ 2
= 0.7071 N 0
The number decaying in this first half of the first half-life is N0 − 0.7071 N0 = 0.2929 N0.
The number remaining after time T1/2 is 0.500 N0, so the number decaying in the second half
of the first half-life is 0.7071 N0 − 0.500 N0 = 0.2071 N0.
The ratio required is then 0.2929 N0/0.2071 N0 = 1.41
549
550
P44.26
Chapter 44
(a)
dN 2
= rate of change of N 2
dt
= rate of production of N 2 − rate of decay of N 2
= rate of decay of N1 − rate of decay of N 2
= λ1 N1 − λ2 N 2
(b)
From the trial solution
N λ
N 2 (t ) = 10 1 ( e− λ2t − e− λ1t )
λ1 − λ2
∴
N λ
dN 2
= 10 1 ( − λ2 e− λ2t + λ1e− λ1t )
dt
λ1 − λ2
∴
N λ
dN 2
+ λ2 N 2 = 10 1 ( − λ2 e− λ2t + λ1e− λ1t + λ2 e− λ2t − λ2 e− λ1t )
dt
λ1 − λ2
=
(1)
N10 λ1
(λ − λ2 ) e− λ1t
λ1 − λ2 1
= λ1 N1
So
The functions to be plotted are
− 0.223 6 min −1 )t
N t = 1 000 e (
1
Decay of 218Po and 214Pb
()
1 200
− 0.223 6 min −1 )t
− 0.025 9 min −1 )t ⎤
−e (
N 2 (t ) = 1130.8 ⎡⎢e (
⎣
⎦⎥
From the graph: t m ≈ 10.9 min
Number of nuclei
(c)
dN 2
= λ1 N1 − λ2 N 2 as required.
dt
1 000
Po
Pb
800
600
400
200
0
0
10
20
time (min)
30
FIG. P44.26(c)
(d)
From (1),
ln ( λ1 λ2 )
dN 2
λ
.
= 0 if λ2 e− λ2t = λ1e− λ1t . ∴ e( λ1 − λ2 )t = 1 . Thus, t = t m =
dt
λ2
λ1 − λ2
With λ1 = 0.223 6 min −1 , λ2 = 0.025 9 min −1 , this formula gives t m = 10.9 min ,
in agreement with the result of part (c).
P44.27
We have all this information: N x ( 0 ) = 2.50 N y ( 0 )
N x (3d ) = 4.20 N y (3d )
N x ( 0 ) e− λx 3d = 4.20 N y ( 0 ) e
− λy 3d
= 4.20
N x ( 0 ) − λy 3d
e
2.50
2.5 3dλy
e
4.2
⎛ 2.5 ⎞
+ 3dλ y
3dλ x = ln ⎜
⎝ 4.2 ⎟⎠
e3dλx =
3d
0.693
0.693
⎛ 2.5 ⎞
+ 3d
= 0.781
= ln ⎜
⎝ 4.2 ⎟⎠
T1 2 x
1.60 d
T1 2 x = 2.66 d
40
Nuclear Structure
Section 44.5
P44.28
A gamma ray has zero charge and it contains no protons or neutrons. So for a gamma
ray Z = 0 and A = 0. Keeping the total values of Z and A for the system conserved then
requires Z = 28 and A = 65 for X. With this atomic number it must be nickel, and the
*
nucleus must be in an exited state, so it is 65
28 Ni .
(b)
α = 42 He
so for X
for Pb
(d)
(e)
P44.29
The Decay Processes
(a)
(c)
Z=2
has
and
we require
and
A=4
Z = 84 − 2 = 82
A = 215 − 4 = 211, X =
211
82
0
−1
e
1
1
H (or p). Note that this process is a nuclear reaction, rather than radioactive decay. We
can solve it from the same principles, which are fundamentally conservation of charge and
conservation of baryon number.
Q = ( M U-238 − M Th-234 − M He-4 ) ( 931.5 MeV u )
⎛ 0.021 0 g ⎞
NC = ⎜
(6.02 × 10 23 moleculees mol )
⎝ 12.0 g mol ⎟⎠
(N
C
Pb
A positron e + = 01 e has charge the same as a nucleus with Z = 1. A neutrino 00ν has
no charge. Neither contains any protons or neutrons. So X must have by conservation
55
Co.
Z = 26 + 1 = 27. It is Co. And A = 55 + 0 = 55. It is 27
Similar reasoning about balancing the sums of Z and A across the reaction reveals:
Q = ( 238.050 783 − 234.043 596 − 4.002 603) u ( 931.5 MeeV u ) = 4.27 MeV
P44.30
551
= 1.05 × 10 21 carbon atoms ) of which 1 in 7.70 × 1011 is a 14C atom
( N 0 )C-14 = 1.37 × 10 9 ,
λC-14 =
ln 2
= 1.21 × 10 −4 yr −1 = 3.83 × 10 −12 s −1
5 730 yr
R = λ N = λ N 0 e− λ t
At t = 0,
⎡ 7 (86 400 s ) ⎤
R0 = λ N 0 = ( 3.83 × 10 −12 s −1 ) (1.37 × 10 9 ) ⎢
⎥
⎣ 1 week ⎦
= 3.17 × 10 3 decays week
At time t,
R=
Taking logarithms,
ln
837
= 951 decays week
0.88
R
= −λ t
R0
t=
so
t=
−1 ⎛ R ⎞
ln
λ ⎜⎝ R0 ⎟⎠
⎛
951 ⎞
−1
ln ⎜
= 9.96 × 10 3 yr
−4
−1
⎝ 3.17 × 10 3 ⎟⎠
1.21 × 10 yr
552
Chapter 44
*P44.31 (a)
(b)
The decay constant is λ = ln 2/10 h = 0.0693/h.
The number of parent nuclei is given by 106 e−0.0693 t where t is in hours.
The number of daughter nuclei is equal to the number of missing parent nuclei,
Nd = 106 − 106 e−0.0693t Nd = 106(1 − e−0.0693t) where t is in hours .
The number of daughter nuclei starts from zero at t = 0. The number of stable product
nuclei always increases with time and asymptotically approaches 1.00 × 106 as t increases
without limit.
dN d
1
= 10 6 (0 + 0.0693 e−0.0693t ) = 6.93 × 10 4 e−0.0693t .
Its rate of increase is
dt
h
The number of daughter nuclei first increases most rapidly, at 6.93 × 104/h, and then more
and more slowly. Its rate of change approaches zero in the far future.
P44.32
3
1
H nucleus → 23 He nucleus + e − + ν
−
−
3
3
becomes
1 H nucleus + e → 2 He nucleus + 2e + ν
Ignoring the slight difference in ionization energies,
3
3
we have
1 H atom → 2 He atom + ν
3.016 049 u = 3.016 029 u + 0 +
Q
c2
Q = (3.016 049 u − 3.016 029 u )( 931.5 MeV u ) = 0.018 6 MeV = 18.6 keV
P44.33
(a)
(b)
(c)
e− + p → n + ν
15
For nuclei,
O + e − → 15 N + ν
Add seven electrons to both sides to obtain
15
8
O atom → 157 N atom + ν
From the table of isotopic masses in the chapter text,
∆m = 15.003 065 u − 15.000 109 u = 0.002 956 u
m ( 15 O ) = m ( 15 N ) +
Q
c2
Q = ( 931.5 MeV u )( 0.002 956 u ) = 2.75 MeV
P44.34
(a)
For e+ decay,
Q = ( M X − M Y − 2 me ) c 2 = ⎡⎣39.962 591 u − 39.963 999 u − 2 ( 0.000 549 u )⎤⎦ ( 931.5 MeV u )
Q = −2.33 MeV
Since Q < 0, the decay cannot occur spontaneously.
(b)
For alpha decay,
Q = ( M X − M α − M Y ) c 2 = [ 97.905 287 u − 4.002 603 u − 93.905 088 u ]( 931.5 MeV u )
Q = −2.24 MeV
Since Q < 0, the decay cannot occur spontaneously.
(c)
For alpha decay,
Q = ( M X − M α − M Y ) c 2 = [143.910 083 u − 4.002 603 u − 139.905 434 u ]( 931.5 MeV u )
Q = 1.91 MeV
Since Q > 0, the decay can occur spontaneously.
Nuclear Structure
Section 44.6
Natural Radioactivity
P44.35
FIG. P44.35
P44.36
(a)
Let N be the number of
238
U nuclei and N ′ be
206
Pb nuclei.
Then N = N 0 e− λ t and N 0 = N + N ′ so N = ( N + N ′ ) e− λ t or eλ t = 1 +
Taking logarithms,
N ′⎞
⎛
λ t = ln ⎜ 1 + ⎟ where
⎝
N⎠
Thus,
N ′⎞
⎛ T1 2 ⎞ ⎛
t=⎜
ln ⎜ 1 + ⎟
⎝ ln 2 ⎟⎠ ⎝
N⎠
If
N
= 1.164 for the
N′
238
U→
206
λ=
N′
.
N
ln 2
T1 2
Pb chain with T1 2 = 4.47 × 10 9 yr, the age is:
⎛ 4.47 × 10 9 yr ⎞ ⎛
1 ⎞
9
t=⎜
⎟⎠ ln ⎝ 1 + 1.164 ⎠ = 4.00 × 10 yr
⎝
ln 2
(b)
From above, eλ t = 1 +
N′
N
N
e− λ t
. Solving for
gives
.
=
N
N′
N ′ 1 − e− λ t
With t = 4.00 × 10 9 yr and T1 2 = 7.04 × 108 yr for the
235
U→
207
Pb chain,
⎛ ln 2 ⎞
( ln 2 ) ( 4.00 × 10 9 yr )
N
= 0.019 9
λt = ⎜
t
=
= 3.938 and
⎟
8
T
N
7
.
04
×
10
yr
′
⎝ 12 ⎠
With t = 4.00 × 10 9 yr and T1 2 = 1.41 × 1010 yr for the
λt =
( ln 2 ) ( 4.00 × 10 9 yr )
1.41 × 1010 yr
232
Th →
= 0.196 6 and
208
Pb chain,
N
= 4.60
N′
553
554
P44.37
Chapter 44
(a)
⎛ 4.00 × 10 −12 Ci ⎞ ⎛ 3.70 × 1010 Bq ⎞ ⎛ 1.00 × 10 3 L ⎞
3
4.00 pCi L = ⎜
⎟⎠ ⎜⎝
⎟⎠ ⎜⎝
⎟⎠ = 148 Bq m
⎝
1 Ci
1 m3
1L
(b)
N=
(c)
1 mol
⎛
⎞ ⎛ 222 g ⎞
mass = ( 7.05 × 10 7 atoms m 3 ) ⎜
= 2.60 × 10 −14 g m 3
⎝ 6.02 × 10 23 atoms ⎟⎠ ⎜⎝ 1 mol ⎟⎠
R
⎛ T1 2 ⎞
⎛ 3.82 d ⎞ ⎛ 86 400 s ⎞
= 7.05 × 10 7 atoms m 3
= R⎜
= (148 Bq m 3 ) ⎜
⎟
⎝ ln2 ⎟⎠ ⎜⎝ 1 d ⎟⎠
⎝ ln 2 ⎠
λ
Since air has a density of 1.20 kg m 3, the fraction consisting of radon is
2.60 × 10 −14 g m 3
= 2.17 × 10 −17
fraction =
1 200 g m 3
.
P44.38
−( ln 2 )t T1 2
Number remaining:
N = N0e
Fraction remaining:
N
−( ln 2 )t T1 2
= e− λ t = e
N0
(a)
With T1 2 = 3.82 d and t = 7.00 d,
N
= e−( ln 2)( 7.00 ) (3.82) = 0.281
N0
(b)
When t = 1.00 yr = 365.25 d,
N
= e−( ln 2)(365.25) (3.82) = 1.65 × 10 −29
N0
(c)
Radon is continuously created as one daughter in the series of decays starting from the
long-lived isotope 238 U.
Section 44.7
P44.39
(a)
(b)
(c)
Nuclear Reactions
For X,
and
A = 24 + 1 − 4 = 21
Z = 12 + 0 − 2 = 10,
A = 235 + 1 − 90 − 2 = 144
Z = 92 + 0 − 38 − 0 = 54,
and
so X is
21
10
so X is
144
54
Ne
Xe
A= 2−2= 0
and
Z = 2 − 1 = +1,
As it is ejected, so is a neutrino:
so X must be a positron.
X = 01 e + and X ′ = 00ν
Nuclear Structure
P44.40
(a)
555
Add two electrons to both sides of the reaction to have it in neutral-atom terms:
Q = ∆mc 2 = ⎡4 M 1 H − M 4 He ⎤ c 2
4 11 H atom → 42 He atom + Q
⎣
⎦
1
2
⎛ 1.60 × 10 −13 J ⎞
Q = ⎡⎣4 (1.007 825 u ) − 4.002 603 u ⎤⎦ ( 931.5 MeV u ) ⎜
= 4.28 × 10 −12 J
⎝ 1 MeV ⎟⎠
1.99 × 10 30 kg
= 1.19 × 10 57 attoms = 1.19 × 10 57 protons
1.67 × 10 −27 kg atom
(b)
N=
(c)
The energy that could be created by this many protons in this reaction is:
(1.19 × 10
P=
P44.41
197
79
(a)
(b)
E
∆t
57
⎛ 4.28 × 10 −12 J ⎞
protons ) ⎜
= 1.27 × 10 45 J
⎝ 4 protons ⎟⎠
so
∆t =
E
P
=
1.27 × 10 45 J
= 3.31 × 1018 s = 105 billion years
3.85 × 10 26 W
*
198
Au + 01 n → 198
79 Au → 80 Hg +
0
−1
e +ν
Consider adding 79 electrons:
197
79
Au atom + 01 n → 198
80 Hg atom + ν + Q
Q = ⎡⎣ M 197 Au + mn − M 198 Hg ⎤⎦ c 2
Q = [196.966 552 + 1.008 665 − 197.966 752 ] u ( 931.5 MeV u ) = 7.89 MeV
P44.42
Neglect recoil of product nucleus (i.e., do not require momentum conservation for the system of
colliding particles). The energy balance gives K emerging = K incident + Q. To find Q:
Q = ⎡⎣( M H + M Al ) − ( M Si + mn )⎤⎦ c 2
Q = ⎡⎣(1.007 825 + 26.981 539 ) − ( 26.986 705 + 1.008 665)⎤⎦ u ( 931.5 MeV u ) = −5.59 MeV
Thus, K emerging = 6.61 MeV − 5.59 MeV = 1.02 MeV .
P44.43
9
4
Be + 1.665 MeV → 48 Be + 01 n, so M 8 Be = M 9 Be −
4
M 8 Be = 9.012 182 u −
4
9
4
( −1.665 MeV )
931.5 MeV u
Q
− mn
c2
− 1.008 665 u = 8.005 3 u
Be + 01 n → 104 Be + 6.812 MeV, so M 10 Be = M 9 Be + mn −
4
M 10 Be = 9.012 182 u + 1.008 665 u −
4
P44.44
4
(a)
10
5
B + 42 He → 136 C + 11 H
The product nucleus is
(b)
13
6
13
6
C .
C + 11 H → 105 B + 42 He
The product nucleus is
10
5
B .
4
Q
c2
6.812 MeV
= 10.013 5 u
931.5 MeV u
556
Chapter 44
Section 44.8
P44.45
Nuclear Magnetic Resonance and Magnetic Resonance Imagining
−27
J T ) (1.00 T )
2 µ B 2 (1.913 5) (5.05 × 10
=
= 29.2 MHz
h
6.626 × 10 −34 J ⋅ s
(a)
fn =
(b)
fp =
(c)
In the Earth’s magnetic field,
fp =
2 ( 2.792 8 ) ( 5.05 × 10 −27 J T ) (1.00 T )
6.626 × 10 −34 J ⋅ s
2 ( 2.792 8 ) ( 5.05 × 10 −27 ) ( 50.0 × 10 −6 )
6.626 × 10 −34
= 42.6 MHz
= 2.13 kHz
P44.46
FIG. P44.46
Additional Problems
P44.47
(a)
Q = ⎡⎣ M 9 Be + M 4 He − M 12 C − mn ⎤⎦ c 2
Q = [ 9.012 182 u + 4.002 603 u − 12.000 000 u − 1.008 665 u ]( 931.5 MeV u ) = 5.70 MeV
(b)
Q = ⎡⎣2 M 2 H − M 3 He − mn ⎤⎦
Q = ⎡⎣2 ( 2.014 102 ) − 3.016 029 − 1.008 665⎤⎦ u ( 931.5 MeV u ) = 3.27 MeV ( exothermic )
P44.48
(a)
With mn and vn as the mass and speed of the neutrons, Equation 9.24 of the text becomes,
⎛ 2 mn ⎞
vn , and
after making appropriate notational changes, for the two collisions v1 = ⎜
⎝ mn + m1 ⎟⎠
⎛ 2 mn ⎞
v2 = ⎜
vn
⎝ mn + m2 ⎟⎠
∴ ( mn + m2 ) v2 = ( mn + m1 ) v1 = 2 mn vn
∴ mn ( v2 − v1 ) = m1v1 − m2 v2
∴ mn =
(b)
mn =
m1v1 − m2 v2
v2 − v1
(1 u ) (3.30 × 10 7 m s ) − (14 u ) ( 4.70 × 106 m s )
4.770 × 10 6 m s − 3.30 × 10 7 m s
= 1.16 u
Nuclear Structure
P44.49
(a)
557
At threshold, the particles have no kinetic energy relative to each other. That is, they move
like two particles that have suffered a perfectly inelastic collision. Therefore, in order
to calculate the reaction threshold energy, we can use the results of a perfectly inelastic
collision. Initially, the projectile M a moves with velocity va while the target M X is at rest.
We have from momentum conservation for the projectile-target system:
M a va = ( M a + M X ) vc
The initial energy is: Ei =
1
M a va2
2
The final kinetic energy is:
2
Ef =
⎡ M v ⎤ ⎡ Ma ⎤
1
1
( M a + M X ) vc2 = ( M a + M X ) ⎢ a a ⎥ = ⎢
⎥ Ei
2
2
⎣ Ma + M X ⎦ ⎣ Ma + M X ⎦
From this, we see that Ef is always less than Ei and the change in energy, E f − Ei , is given by
⎡ Ma
⎤
⎡ MX ⎤
E f − Ei = ⎢
− 1⎥ Ei = − ⎢
⎥ Ei
⎣ Ma + M X
⎦
⎣ Ma + M X ⎦
This loss of kinetic energy in the isolated system corresponds to an increase in mass-energy
during the reaction. Thus, the absolute value of this kinetic energy change is equal to −Q
(remember that Q is negative in an endothermic reaction). The initial kinetic energy Ei is
the threshold energy Eth.
(b)
Therefore,
⎡ MX ⎤
−Q = ⎢
⎥ Eth
⎣ Ma + M X ⎦
or
⎡ M + Ma ⎤
⎡ Ma ⎤
Eth = −Q ⎢ X
⎥ = −Q ⎢1 + M ⎥
M
⎣
⎦
⎣
X
X ⎦
First, calculate the Q-value for the reaction:
Q = [ M N-14 + M He-4 − M O-17 − M H-1 ] c 2
Q = [14.003 074 + 4.002 603 − 16.999 132 − 1.007 825] u ( 9331.5 MeV u ) = −1.19 MeV
⎡ M + Ma ⎤
⎡
4.002 603 ⎤
= 1.53 MeV .
Then, Eth = −Q ⎢ X
⎥ = − ( −1.19 MeV ) ⎢1 +
4.003 074 ⎥⎦
M
1
⎣
⎣
⎦
X
*P44.50 (a)
(b)
The system of a separated proton and electron puts out energy 13.606 eV to become
a hydrogen atom in its ground state. This decrease in its rest energy appears also as a
decrease in mass: the mass is smaller .
∆m =
⎛ 1.6 × 10 −19 J ⎞
13.6 eV
E
=
= 2.42 × 10 −35 kg
2
2
c
(3 × 108 m s) ⎜⎝ 1 eV ⎟⎠
⎛
⎞
1u
= 2.42 × 10 −35 kg ⎜
= 1.46 × 10 −8 u
⎝ 1.66 × 10 −27 kg ⎟⎠
(c)
(d)
1.46 × 10 −8 u
= 1.45 × 10 −8 = 1.45 × 10 −6%
1.007 825 u
The textbook table lists 1.007 825 u as the atomic mass of hydrogen. This correction of
0.000 000 01 u is on the order of 100 times too small to affect the values listed.
558
Chapter 44
*P44.51 The problem statement can be: Find the reaction energy (Q value) of the reaction
10
4
13
1
5 B + 2 He → 6 C + 1 H.
Solving gives Q = (14.015 54 − 14.011 18) u c2(931.5 MeV/u c2) = 4.06 MeV for the energy
released by the reaction as it is converted from rest energy into other forms.
P44.52
(a)
N0 =
(b)
λ=
1.00 kg
mass
=
= 2.52 × 10 24
mass per atom ( 239.05 u ) (1.66 × 10 −27 kg u )
ln 2
ln 2
=
= 9.106 × 10 −13 s −1
T1 2 ( 2.412 × 10 4 yr ) (3.156 × 10 7 s yr )
R0 = λ N 0 = ( 9.106 × 10 −13 s −1 ) ( 2.52 × 10 24 ) = 2.29 × 1012 Bq
(c)
R = R0 e− λ t , so t =
t=
P44.53
(a)
57
27
−1 ⎛ R ⎞ 1 ⎛ R0 ⎞
ln
= ln ⎜ ⎟
λ ⎜⎝ R0 ⎟⎠ λ ⎝ R ⎠
⎛ 2.29 × 1012 Bq ⎞
1
ln
= 3.38 × 1013
9.106 × 10 −13 s −1 ⎜⎝ 0.100 Bq ⎟⎠
Co →
57
26
Fe +
0
+1
1 yr
⎛
⎞
s⎜
= 1.07 × 10 6 yr
⎝ 3.156 × 10 7 s ⎟⎠
e + 00ν
The Q-value for this positron emission is Q = ⎡⎣ M 57 Co − M 57 Fe − 2 me ⎤⎦ c 2 .
Q = ⎡⎣56.936 296 − 56.935 399 − 2 ( 0.000 549 )⎤⎦ u ( 931.5 MeV u ) = −0.187 MeV
Since Q < 0, this reaction cannot spontaneously occur .
(b)
14
6
C → 147 N +
0
−1
e + 00ν
The Q-value for this e− decay is Q = ⎡⎣ M 14 C − M 14 N ⎤⎦ c 2 .
Q = [14.003 242 − 14.003 074 ] u ( 931.5 MeV u ) = 0.156 MeV = 156 keV
Since Q > 0, the decay can spontaneously occur .
P44.54
(c)
The energy released in the reaction of (b) is shared by the electron and neutrino. Thus,
K e can range from zero to 156 keV .
(a)
r = r0 A1 3 = 1.20 × 10 −15 A1 3 m
r = 2.75 × 10 −15 m
When A = 12,
(b)
9
2
2
−19
k ( Z − 1) e2 (8.99 × 10 N ⋅ m C ) ( Z − 1) (1.60 × 10 C )
F= e
=
r2
r2
When Z = 6
(c)
(d)
and
r = 2.75 × 10 −15 m,
kqq
k ( Z − 1) e2 (8.99 × 10
U= e 1 2 = e
=
r
r
9
2
F = 152 N
) ( Z − 1) (1.6 × 10 )
−19 2
r
−15
When Z = 6
and
r = 2.75 × 10
A = 238;
Z = 92,
r = 7.44 × 10 −15 m
and
U = 2.82 × 10 −12 J = 17.6 MeV
m,
U = 4.19 × 10 −13 J = 2.62 MeV
F = 379 N
Nuclear Structure
P44.55
(a)
Because the reaction p → n + e + + ν would violate the law of conservation of energy
m p = 1.007 276 u
Note that
mn = 1.008 665 u
me + = 5.49 × 10 −4 u
mn + me + > m p
(b)
The required energy can come from the electrostaatic repulsion of protons in the parent
nucleus.
(c)
Add seven electrons to both sides of the reaction for nuclei 137 N → 136 C + e + + ν
to obtain the reaction for neutral atoms 137 N atom → 136 C atom + e + + e − + ν
Q = c 2 ⎡⎣ m ( 13 N ) − m ( 13 C ) − me+ − me− − mν ⎤⎦
Q = ( 931.5 MeV u ) ⎡⎣13.005 739 − 13.003 355 − 2 (5.49 × 10 −4 ) − 0 ⎤⎦ u
Q = ( 931.5 MeV u ) (1.286 × 10 −3 u ) = 1.20 MeV
P44.56
(a)
A least-square fit to the graph yields:
λ = −slope = − ( −0.250 h −1 ) = 0.250 h −1
and
(b)
ln ( cpm ) t = 0 = intercept = 8.30
⎛ 1h ⎞
= 4.17 × 10 −3 min −1
λ = 0.250 h −1 ⎜
⎝ 60.0 min ⎟⎠
T1 2 =
ln 2
ln 2
=
λ
4.17 × 10 −3 min −1
= 166 min = 2.77 h
(c)
P44.57
FIG. P44.56
From (a),
intercept = ln ( cpm )0 = 8.30
Thus,
( cpm )0 = e8.30
counts min = 4.02 × 10 3 counts min
R0 1 ( cpm )0
4.02 × 10 3 counts min
=
=
= 9.65 × 10 6 atoms
λ λ Eff
(4.17 × 10 −3 min −1 ) (0.100 )
(d)
N0 =
(a)
One liter of milk contains this many 40 K nuclei:
⎛ 6.02 × 10 23 nuclei mol ⎞ ⎛ 0.011 7 ⎞
18
N = ( 2.00 g ) ⎜
⎜
⎟ = 3.60 × 10 nuclei
⎝
⎠⎟ ⎝ 100 ⎠
39.1 g mol
λ=
ln 2
ln 2
1 yr
⎛
⎞
−17 −1
=
⎜
⎟ = 1.72 × 10 s
T1 2 1.28 × 10 9 yr ⎝ 3.156 × 10 7 s ⎠
R = λ N = (1.72 × 10 −17 s −1 ) (3.60 × 1018 ) = 61.8 Bq
(b)
For the iodine, R = R0 e− λ t
t=
with
λ=
1 ⎛ R0 ⎞ 8.04 d ⎛ 2 000 ⎞
= 40.3 d
ln ⎜ ⎟ =
ln ⎜
⎝ 61.8 ⎟⎠
λ ⎝ R⎠
ln 2
ln 2
8.04 d
559
560
P44.58
Chapter 44
(a)
If ∆E is the energy difference between the excited and ground states of the nucleus of mass
M, and hf is the energy of the emitted photon, conservation of energy for the
nucleus-photon system gives
∆E = hf + Er
(1)
Where Er is the recoil energy of the nucleus, which can be expressed as
Er =
M v2 ( M v)
=
2
2M
2
(2)
Since system momentum must also be conserved, we have
Mv =
(b)
hf
c
( hf )2
Hence, Er can be expresses as
Er =
When
hf << Mc 2
we can make the approximation that
hf ≈ ∆E
so
Er ≈
Er =
( ∆E )2
2 Mc 2
( ∆E )2
2 Mc 2
where
2 Mc 2
(3)
∆E = 0.014 4 MeV
and
Mc 2 = ( 57 u ) ( 931.5 MeV u ) = 5.31 × 10 4 MeV
Therefore,
Er =
(1.44 × 10 MeV ) =
2 ( 5.31 × 10 MeV )
2
−2
P44.59
We have
N 235 = N 0 ,235 e− λ235t
and
N 238 = N 0 ,238 e− λ238t
N 235
( −( ln 2)t Th ,235 +( ln 2)t Th , 238 )
= 0.007 25 = e
N 238
Taking logarithms,
⎛
⎞
ln 2
ln 2
−4.93 = ⎜ −
+
t
9
9
⎝ 0.704 × 10 yr 4.47 × 10 yr ⎟⎠
or
⎞
⎛
1
1
−4.93 = ⎜ −
+
( ln 2 ) t
⎝ 0.704 × 10 9 yr 4.47 × 10 9 yr ⎟⎠
t=
−4.93
= 5.94 × 10 9 yr
( −1.20 × 10 −9 yr −1 ) ln 2
4
1.94 × 10 −3 eV
Nuclear Structure
P44.60
(a)
For cobalt-56,
λ=
ln 2
ln 2 ⎛ 365.25 d ⎞
= 3.28 yrr −1
=
T1 2 77.1 d ⎜⎝ 1 yr ⎟⎠
The elapsed time from July 1054 to July 2007 is 953 yr.
R = R0 e− λ t
implies
(b)
R
−( 3.28 yr −1 )( 953 yr )
= e− λ t = e
= e−3 129 = e−( ln10 )1 359 = ~10 −1 359
R0
For carbon-14,
λ=
ln 2
= 1.21 × 10 −4 yr −1
5 730 yr
R
− 1.21×10 −4 yr −1 )( 953 yr )
= e− λ t = e (
= e−0.115 = 0.891
R0
P44.61
E = − µ ⋅ B so the energies are E1 = + µ B and E2 = − µ B
µ = 2.792 8µn and µn = 5.05 × 10 −27 J T
∆E = 2 µ B = 2 ( 2.792 8 ) ( 5.05 × 10 −27 J T ) (12.5 T ) = 3.53 × 10 −25 J = 2.20 × 10 −6 eV
P44.62
R = R0 exp ( − λ t ) lets us write ln R = ln R0 − λ t
which is the equation of a straight line with
slope = λ . The logarithmic plot shown in
Figure P44.62 is fitted by ln R = 8.44 − 0.262t .
If t is measured in minutes, then decay constant λ is
0.262 per minute. The half−life is
ln 2
ln 2
T1 2 =
=
= 2.64 min .
λ
0.262 min
The reported half−life of 137 Ba is 2.55 min. The
difference reflects experimental uncertainties.
FIG. P44.62
561
562
P44.63
Chapter 44
K=
1
mv 2 ,
2
so
v=
2 ( 0.040 0 eV ) (1.60 × 10 −19 J eV )
2K
=
m
The time for the trip is t =
1.67 × 10 −27 kg
= 2.77 × 10 3 m s
x
1.00 × 10 4 m
=
= 3.61 s.
v 2.77 × 10 3 m s
The number of neutrons finishing the trip is given by N = N 0 e− λ t .
The fraction decaying is 1 −
P44.64
N
−( ln 2 )t T1 2
= 1− e
= 1 − e−(ln 2 )(3.61 s 624 s) = 0.004 00 = 0.400% .
N0
Ze
.
( 4 3) π R 3
Using Gauss’s Law inside the sphere,
For an electric charge density ρ =
E ⋅ 4π r 2 =
( 4 3) π r 3
∈0
Ze
:
( 4 3) π R 3
E=
1 Zer
4π ∈0 R3
(r ≤ R )
E=
1 Ze
4π ∈0 r 2
(r ≥ R )
∞
U=
We now find the electrostatic energy:
1
∈0 E 2 4π r 2 dr
2
r=0
∫
2
2
R
⬁
⎛ 1 ⎞ Z 2 e2 r 2
⎛ 1 ⎞ Z 2 e2
1
Z 2 e2 ⎡ R 5 1 ⎤
1
2
2
+
∈
=
+
U = ∈0 ∫ ⎜
4
π
r
dr
4
π
r
dr
0
4
6
∫R ⎜⎝ 4π ∈0 ⎟⎠ r
8π ∈0 ⎢⎣ 5 R6 R ⎥⎦
2
4π ∈0 ⎟⎠
R
2
0⎝
=
P44.65
(a)
3 Z 2 e2
20 π ∈0 R
If we assume all the 87 Sr came from
N = N0e
yields
t=
where
N = N Rb-87
and
N 0 = N Sr-87 + N Rb-87
−1 ⎛ N ⎞ T1 2 ⎛ N 0 ⎞
ln
ln ⎜ ⎟
=
λ ⎜⎝ N 0 ⎟⎠ ln 2 ⎝ N ⎠
( 4.75 × 10
10
(b)
Rb,
−λ t
then
t=
87
ln 2
yr )
⎛ 1.82 × 1010 + 1.07 × 10 9 ⎞
9
ln ⎜
⎟⎠ = 3.91 × 10 yr
⎝
1.82 × 1010
It could be no older . The rock could be younger if some 87 Sr were originally present.
We must make some assumption about the original quantity of radioactive material. In
part (a) we assumed that the rock originally contained no strontium.
Nuclear Structure
P44.66
(a)
Tc +
93
43
For the electron capture,
0
−1
e→
93
42
Mo + γ
The disintegration energy is Q = ⎡⎣ M 93 Tc − M 93 Mo ⎤⎦ c 2 .
Q = [ 92.910 2 − 92.906 8] u ( 931.5 MeV u ) = 3.17 MeV > 2.44 MeV
Electron capture is allowed
93
to all specified excited states in 42
Mo.
Tc →
93
43
For positron emission,
93
42
Mo +
0
+1
e+γ
The disintegration energy is Q ′ = ⎡⎣ M 93 Tc − M 93 Mo − 2 me ⎤⎦ c 2 .
Q ′ = ⎡⎣92.910 2 − 92.906 8 − 2 ( 0.000 549 )⎤⎦ u ( 931.5 MeV
V u ) = 2.14 MeV
Positron emission can reach
the 1.35, 1.48, and 2.03 MeV states
but there is insufficient energy to reach
the 2.44 MeV state.
(b)
The daughter nucleus in both forms of
93
decay is 42
Mo .
FIG. P44.66
ANSWERS TO EVEN PROBLEMS
P44.2
(a) 7.89 cm and 8.21 cm
(b) See the solution.
P44.4
(a) 29.5 fm
(b) 5.18 fm
(c) See the solution.
P44.6
(a) 1.90 fm
(b) 7.44 fm
P44.8
6.11 PN toward the other ball
P44.10
They agree with the figure.
P44.12
0.210 MeV greater for
P44.14
(a) 84.1 MeV
repulsion.
P44.16
(a) See the solution.
P44.18
(a) 491 MeV (b) 179%, −53.0%, −24.6%, −1.37%
P44.20
0.200 mCi
P44.22
See the solution.
P44.24
R0T1 2
ln 2
(2
− t1 T1 2
23
Na because it has less proton repulsion
(b) 342 MeV
−2
− t 2 T1 2
(b)
)
(c) The nuclear force of attraction dominates over electrical
R
R
and ; see the solution.
3
6
563
564
Chapter 44
(c) See the solution;10.9 min.
P44.26
(a) See the solution. (b) See the solution.
ln ( λ1 λ2 )
; yes
(d) t m =
λ1 − λ2
P44.28
(a)
P44.30
9.96 × 103 yr
P44.32
3
1
P44.34
(a) cannot occur
P44.36
(a) 4.00 Gyr
P44.38
(a) 0.281
P44.40
(a) 4.28 pJ
P44.42
1.02 MeV
P44.44
(a) 136 C
P44.46
See the solution.
P44.48
(a) See the solution.
P44.50
(a) smaller
P44.52
(a) 2.52 × 10 24
P44.54
(a) 2.75 fm
P44.56
(a) See the solution. (b) 4.17 × 10 −3 min −1 ; 2.77 h
(d) 9.65 × 10 6 atoms
P44.58
(a) See the solution.
P44.60
(a) ~10 −1 359
P44.62
2.64 min
P44.64
See the solution.
P44.66
(a) electron capture to all; positron emission to the 1.35 MeV, 1.48 MeV, and 2.03 MeV states
93
Mo; see the solution.
(b) 42
65
28
Ni*
(b)
H atom →
3
2
211
82
Pb
(c)
55
27
Co
(d)
0
−1
e
(e) 11 H
He atom + ν ; 18.6 keV
(b) cannot occur
(c) can occur
(b) 0.019 9 and 4.60
(b) 1.65 × 10 −29
(c) See the solution.
(b) 1.19 × 10 57 atoms
(c) 105 Gyr
(b) 105 B
(b) 1.16 u
(b) 1.46 × 10−8 u
(b) 2.29 TBq
(b) 152 N
(c) 1.45 × 10−6 %
(d) no
(c) 1.07 Myr
(c) 2.62 MeV
(d) 7.44 fm, 379 N, 17.6 MeV
(c) 4.02 × 10 3 counts min
(b) 1.94 meV
(b) 0.891
45
Applications of Nuclear Physics
CHAPTER OUTLINE
45.1
45.2
45.3
45.4
45.5
45.6
45.7
Interactions Involving Neutrons
Nuclear Fission
Nuclear Reactors
Nuclear Fusion
Radiation Damage
Radiation Detectors
Uses of Radiation
ANSWERS TO QUESTIONS
Q45.1
The hydrogen nuclei in water molecules have mass
similar to that of a neutron, so that they can efficiently
rob a fast-moving neutron of kinetic energy as they
scatter it. Once the neutron is slowed down, a hydrogen
nucleus can absorb it in the reaction n + 11 H → 21 H.
Q45.2
The excitation energy comes from the binding energy of
the extra nucleon.
*Q45.3 Answer (c). The function of the moderator is to slow
down the neutrons released by one fission so that they
can efficiently cause more fissions.
Q45.4
The advantage of a fission reaction is that it can generate much more electrical energy per
gram of fuel compared to fossil fuels. Also, fission reactors do not emit greenhouse gases as
combustion byproducts like fossil fuels—the only necessary environmental discharge is heat.
The cost involved in producing fissile material is comparable to the cost of pumping, transporting
and refining fossil fuel.
The disadvantage is that some of the products of a fission reaction are radioactive—and some
of those have long half-lives. The other problem is that there will be a point at which enough
fuel is spent that the fuel rods do not supply power economically and need to be replaced.
The fuel rods are still radioactive after removal. Both the waste and the “spent” fuel rods present
serious health and environmental hazards that can last for tens of thousands of years. Accidents
and sabotage involving nuclear reactors can be very serious, as can accidents and sabotage
involving fossil fuels.
*Q45.5 Answer (b). In m = E/c2 we estimate 5 × 1013 J/9 × 1016 m2/s2 ≈ 6 × 10–4 kg = 0.6 g.
*Q45.6 Answer (d). We compute 235 + 1 – 137 – 96 = 3. All the protons that start out in the uranium
nucleus end up in the fission product nuclei.
Q45.7
The products of fusion reactors are generally not themselves unstable, while fission reactions
result in a chain of reactions which almost all have some unstable products.
Q45.8
For the deuterium nuclei to fuse, they must be close enough to each other for the nuclear forces to
overcome the Coulomb repulsion of the protons—this is why the ion density is a factor. The more
time that the nuclei in a sample spend in close proximity, the more nuclei will fuse—hence the
confinement time is a factor.
Q45.9
Fusion of light nuclei to a heavier nucleus releases energy. Fission of a heavy nucleus to lighter
nuclei releases energy. Both processes are steps towards greater stability on the curve of binding
energy, Figure 44.5. The energy release per nucleon is typically greater for fusion, and this
process is harder to control.
565
566
Chapter 45
Q45.10 Advantages of fusion: high energy yield, no emission of greenhouse gases, fuel very easy to
obtain, reactor cannot go supercritical like a fission reactor, low amounts of radioactive waste.
Disadvantages: requires high energy input to sustain reaction, lithium and helium are scarce,
neutrons released by reaction cause structural damage to reactor housing.
*Q45.11 Answer Q1 > Q2 > Q3 > 0. Because all of the reactions involve 108 nucleons, we can look just at
the change in binding-energy-per-nucleon as shown on the curve of binding energy. The jump
from lithium to carbon is the biggest jump, and next the jump from A = 27 to A = 54, which is near
the peak of the curve. The step up for fission from A = 108 to A = 54 is smallest. Both of the fusion
reactions described and the fission reaction put out energy, so Q is positive for all. Imagine turning
the curve of binding energy upside down so that it bends down like a cross-section of a bathtub.
On such a curve of total energy per nucleon versus mass number it is easy to identify the fusion
of small nuclei, the fission of large nuclei, and even the alpha decay of uranium, as exoenergetic
processes. The most stable nucleus is at the drain of the bathtub, with minimum energy.
*Q45.12 Answer (d). The particles lose energy by collisions with nuclei in the bubble chamber to make
their speed and their cyclotron radii r = mv/qB decrease.
*Q45.13 Answer (d) only. The Geiger counter responds to an individual particle with a pulse whose size is
determined by the tube power supply.
Q45.14 For each additional dynode, a larger applied voltage is needed, and hence a larger output from
a power supply—“infinite” amplification would not be practical. Nor would it be desirable: the
goal is to connect the tube output to a simple counter, so a massive pulse amplitude is not needed.
If you made the detector sensitive to weaker and weaker signals, you would make it more and
more sensitive to background noise.
*Q45.15 Answer (b). The cyclotron radius is given by r = mv/qB = 2 12 m 2 v 2 / qB = 2 mK / qB. With the
same K, | q |, and B for both particles, the electron with much smaller mass has smaller radius and
is deflected more.
*Q45.16 Answer b > c > a > d. Dose (a) is 1 rem. Dose (b) is 10 rem. Doses (c) and (d) are 4 to 5 rem.
If we assume that (a) and (b) as well as (c) were whole-body doses to many kilograms of tissue,
we find the ranking stated.
Q45.17 Sometimes the references are indeed oblique. Some can serve for more than one form of energy
or mode of transfer. Here is one list:
kinetic: ocean currents
rotational kinetic: Earth turning
gravitational: water lifted up; you on the perilous bridge
elastic: Elastic energy is necessary for sound, listed below.
Vibrational energy could be separately exemplified by the swaying bridge of land.
internal: lava in and from infernal volcanoes; or in forging a chain
chemical: corrosive smoke
sound: thunder
electrical transmission: lightning
electromagnetic radiation: heavens blazing; lightning
atomic electronic: In the blazing heavens, stars have different colors because of different
predominant energy losses by atoms at their surfaces.
nuclear: The blaze of the heavens is produced by nuclear reactions in the cores of stars.
Remarkably, the word “energy” in this translation is an anachronism. Goethe wrote the song a
few years before Thomas Young coined the term.
Applications of Nuclear Physics
567
SOLUTIONS TO PROBLEMS
Section 45.1
Interactions Involving Neutrons
Section 45.2
Nuclear Fission
P45.1
The energy is
⎞⎛ M ⎞
⎛
⎞ ⎛ 1 U-235 nuccleus ⎞ ⎛
235 g
1 eV
3.30 × 1010 J ⎜
⎟⎜
⎟⎜
⎟
⎟⎜
−19
23
⎝ 1.60 × 10 J ⎠ ⎝ 208 MeV ⎠ ⎝ 6.02 × 10 nucleus ⎠ ⎝ 10 6 ⎠
= 0.387 g of U-235
P45.2
∆m = ( mn + M U ) − ( M Zr + M Te + 3mn )
∆m = (1.008 665 u + 235.043 923 u ) − ( 97.912 7 u + 134.916 5 u + 3 (1.008 665 u ))
∆m = 0.197 39 u = 3.28 × 10 −28 kg
P45.3
Q = ∆mc 2 = 2.95 × 10 −11 J = 184 MeV
Three different fission reactions are possible:
1
0
P45.4
so
n+
235
92
U→
90
38
1
Sr + 143
54 Xe + 3 0 n
143
54
Xe
1
0
n+
235
92
144
1
U → 90
38 Sr + 54 Xe + 2 0 n
144
54
Xe
1
0
n+
235
92
U→
142
54
Xe
90
38
Sr +
142
54
Xe + 4 01 n
Q = ( ∆m ) c 2 = [ mn + M U235 − M Ba141 − M Kr92 − 3mn ] c 2
(a)
∆m = ⎡⎣(1.008 665 + 235.043 923) − (140.914 4 + 91.926 2 + 3 × 1.008 665) ⎤⎦ u = 0.185 993 u
Q = ( 0.185 993 u ) ( 931.5 MeV u ) = 173 MeV
f =
(b)
P45.5
P45.6
P45.7
1
0
n+
∆m 0.185993 u
=
= 7.88 × 10 −4 = 0.078 8%
236.05 u
mi
Th →
232
90
Th →
233
90
233
91
Pa + e − + ν
233
91
Pa →
233
92
U + e− + ν
(a)
The initial mass is 1.007 825 u + 11.009 306 u = 12.017 131 u. The final mass is
3 ( 4.002 603 u ) = 12.007 809 u. The rest mass annihilated is ∆m = 0.009 322 u. The
931.5 MeV ⎞
energy created is Q = ∆mc 2 = 0.009 322 u ⎛
= 8.68 MeV .
⎝
⎠
1u
(b)
The proton and the boron nucleus have positive charges. The colliding particles must have
enough kinetic energy to approach very closely in spite of their electric repulsion.
The available energy to do work is 0.200 times the energy content of the fuel.
(1.00 kg fuel) (0.034 0
235
⎛ ( 208 ) (1.60 × 10 −13 J ) ⎞
⎛ 1 000 g ⎞ ⎛ 1 mol ⎞
23
⎟
U fuel) ⎜
⎟⎜
⎟ ( 6.02 × 10 mol) ⎜⎜
⎟
fission
⎝ 1 kg ⎠ ⎝ 235 g ⎠
⎝
⎠
= 2.90 × 1012 J
(2.90 × 10 J) (0.200) = 5.80 × 10
12
∆r = 5.80 × 10 6 m = 5.80 Mm
11
J = (1.00 × 10 5 N ) ∆r
568
P45.8
Chapter 45
If the electrical power output of 1 000 MW is 40.0% of the power derived from fission reactions,
the power output of the fission process is
1 000 MW
= ( 2.50 × 10 9 J s ) (8.64 × 10 4 s d ) = 2.16 × 1014 J d
0.400
The number of fissions per day is
( 2.16 × 10
14
1 fission ⎞ ⎛
1 eV
⎞ = 6.74 × 10 24 d −1
J d)⎛
⎝ 200 × 10 6 eV ⎠ ⎝ 1.60 × 10 −19 J ⎠
This also is the number of
(6.74 × 10
24
235
U nuclei used, so the mass of
235
U used per day is
⎛
235 g mol
⎞
nuclei d ) ⎜
= 2.63 × 10 3 g d = 2.63 kg d
23
⎝ 6.02 × 10 nuclei mol ⎟⎠
In contrast, a coal-burning steam plant producing the same electrical power uses more than
6 × 10 6 kg d of coal.
Section 45.3
P45.9
Nuclear Reactors
Mass of
235
⎛ 10 6 g ⎞
12
U available ≈ ( 0.007 ) (10 9 metric tons) ⎜
⎟ = 7 × 10 g
⎝ 1 metric ton ⎠
⎛ 7 × 1012 g ⎞
number of nuclei ≈ ⎜
(6.02 × 10 23 nuclei mol ) = 1.8 × 1034 nuclei
⎝ 235 g mol ⎟⎠
The energy available from fission (at 208 MeV/event) is
E ≈ (1.8 × 10 34 events ) ( 208 MeV event ) (1.60 × 10 −13 J MeV ) = 6.0 × 10 23 J
This would last for a time interval of
∆t =
E
P
≈
6.0 × 10 23 J
1 yr
⎞ ≈ 3 000 yr
= (8.6 × 1010 s ) ⎛
⎝ 3.16 × 10 7 s ⎠
7.0 × 1012 J s
and
3V
r=⎛ ⎞
⎝ 4π ⎠
V = 3
and
= V1 3
V = 2a 3
and
V
a=⎛ ⎞
⎝ 2⎠
4 3
πr
3
so
A
4π r 2
= 4.84V −1 3
=
V ( 4 3) π r 3
so
A 6 2
= 3 = 6V −1 3
V
13
For a sphere:
V=
(b)
For a cube:
(c)
For a parallelepiped:
(d)
The answers show that the sphere has the smallest surface area for a given volume and the
*P45.10 (a)
13
so
2
2
A ( 2a + 8a )
= 6.30V −1 3
=
V
2a 3
brick has the greatest surface area of the three. Therefore, the sphere has the least leakage
and the parallelepiped has the greatest leakage .
Applications of Nuclear Physics
P45.11
In one minute there are
60.0 s
= 5.00 × 10 4 fissions.
1.20 ms
So the rate increases by a factor of (1.000 25)
50 000
P45.12
569
= 2.68 × 10 5 .
P = 10.0 MW = 1.00 × 10 7 J s. If each decay delivers 1.00 MeV = 1.60 × 10 −13 J, then the
number of decays/s =
*P45.13 (a)
10 7 J/s
= 6.25 × 1019 Bq .
1.6 × 10⫺13 J
Since K = p2/2m, we have
2 mK = 2 m 23 kBT = 3(1.67 × 10 −27 kg)(1.38 × 10 −23 J/K)) 300 K
p=
= 4.55 × 10–24 kg ⋅ m/s
(b)
l = h/p = 6.63 × 10–34 J ⋅ s/4.55 × 10–24 kg ⋅ m/s = 0.146 nm . This size has the same order
of magnitude as an atom’s outer electron cloud, and is vastly larger than a nucleus.
*P45.14 We take the radii of the helium and gold nuclei as 1.20 fm 41/3 = 1.90 fm and 1.20 fm 1971/3 =
6.98 fm. The center to center distance is then 8.89 fm and the electric potential energy is
U = qV =
Section 45.4
P45.15
(a)
ke q1q2 8.99 × 10 9 N ⋅ m 2 2 × 1.6 × 10 −19 C 79 e
= 25.6 MeV
=
C2 8.89 × 10 −15 m
r
Nuclear Fusion
The Q value for the D-T reaction is 17.59 MeV.
Specific energy content in fuel for D-T reaction:
(17.59 MeV ) (1.60 × 10 −13 J MeV )
= 3.39 × 1014 J kg
( 5 u ) (1.66 × 10 −27 kg u )
rDT =
(b)
(3.00 × 10
9
(3.39 × 10
14
J s ) ( 3 600 s hr )
J kg ) (10 −3 kg g )
= 31.9 g h burning of D and T
1
Specific energy content in fuel for D-D reaction: Q = ( 3.27 + 4.03) = 3.65 MeV average
2
of two Q values
( 3.65 MeV ) (1.60 × 10 −13 J MeV )
= 8.80 × 1013 J kg
( 4 u ) (1.66 × 10 −27 kg u )
rDD =
(3.00 × 10
(8.80 × 10
9
13
J s ) ( 3 600 s hr )
J kg ) (10 −3 kg g )
= 122 g h burning of D
570
Chapter 45
*P45.16 (a)
We assume that the nuclei are stationary at closest approach, so that the electrostatic
potential energy equals the total energy E.
Uf =
ke ( Z1e)( Z 2 e)
rmin
(8.99 × 10
E=
9
=E
N ⋅ m 2 C2 ) (1.6 × 10 −19 C ) Z1 Z 2
2
1.00 × 10 −14 m
=
( 2.30 × 10
−14
J ) Z1 Z 2 = (144 keV) Z1Z2
(b)
The energy is proportional to each atomic number.
(c)
Take Z1 = 1 and Z2 = 59 or vice versa. This choice minimizes the product Z1 Z2. If extra
cleverness is allowed, take Z1 = 0 and Z2 = 60: use neutrons as the bombarding particles.
A neutron is a nucleon but not an atomic nucleus.
(d)
For both the D-D and the D-T reactions, Z1 = Z 2 = 1. Thus, the minimum energy required
in both cases is
1 MeV ⎞
E = ( 2.30 × 10 −14 J ) ⎛
= 144 keV
⎝ 1.60 × 10 −13 J ⎠
Section 45.4 in the text gives more accurate values for the critical ignition temperatures, of
about 52 keV for D-D fusion and 6 keV for D-T fusion. The nuclei can fuse by tunneling.
A triton moves more slowly than a deuteron at a given temperature. Then D-T collisions
last longer than D-D collisions and have much greater tunneling probabilities.
P45.17
(a)
13
13
rf = rD + rT = (1.20 × 10 −15 m ) ⎡⎣( 2 ) + ( 3) ⎤⎦ = 3.24 × 10 −15 m
(b)
Uf =
(c)
⎛ mD ⎞
2
Conserving momentum, mD vi = ( mD + mT ) v f or v f = ⎜
⎟ vi = vi
5
m
+
m
⎝ D
T⎠
(d)
K i + Ui = K f + U f :
9
2
2
−19
ke e2 ( 8.99 × 10 N ⋅ m C ) (1.60 × 10 C)
= 7.10 × 10 −14 J = 444 keV
=
rf
3.24 × 10 −15 m
2
⎛ m
⎞
1
1
( mD + mT ) v 2f + Uf = ( mD + mT ) ⎜ D ⎟ vi2 + U f
2
2
⎝ mD + m T ⎠
2
Ki + 0 =
⎛ mD ⎞ ⎛ 1
⎛ mD ⎞
2⎞
Ki + 0 = ⎜
⎟ ⎜ mD vi ⎟ + U f = ⎜
⎟ Ki + U f
⎠
⎝ mD + m T ⎠ ⎝ 2
⎝ mD + m T ⎠
⎛
mD ⎞
⎜1−
⎟ Ki = U f :
⎝ mD + m T ⎠
(e)
⎛ m + mT ⎞ 5
K i = Uf ⎜ D
⎟ = ( 444 keV) = 740 keV
⎝ mT ⎠ 3
The nuclei can fuse by tunneling through thee potential-energy barrier.
Applications of Nuclear Physics
571
3
P45.18
(a)
1 609 m ⎞
V = ( 317 × 10 6 mi 3 ) ⎛
= 1.32 × 1018 m 3
⎝ 1 mi ⎠
mwater = ρV = (10 3 kg m 3 ) (1.32 × 1018 m 3 ) = 1.32 × 10 21 kg
⎛ M H2 ⎞
⎛ 2.016 ⎞
21
20
mH 2 = ⎜
⎟ mH2O = ⎝ 18.015 ⎠ (1.32 × 10 kg ) = 1.48 × 10 kg
M
⎝ H 2O ⎠
mDeuterium = ( 0.030 0%) mH2 = ( 0.030 0 × 10 −2 ) (1.48 × 10 20 kg ) = 4.43 × 1016 kg
The number of deuterium nuclei in this mass is
N=
mDeuterium
4.43 × 1016 kg
=
= 1.33 × 10 43
mDeuteron ( 2.014 u ) (1.66 × 10 −27 kg u )
Since two deuterium nuclei are used per fusion, 21 H + 21 H → 42 He + Q / c 2 , the number of
events is
N
= 6.63 × 10 42.
2
The energy released per event is
Q = ⎡⎣ M 2 H + M 2 H − M 4 He ⎤⎦ c 2 = ⎡⎣ 2 ( 2.014 102 ) − 4.002 603 ⎤⎦ u ( 931.5 MeV u ) = 23.8 MeV
The total energy available is then
⎛ 1.60 × 10 −13 J ⎞
N
= 2.53 × 10 31 J
E = ⎛ ⎞ Q = ( 6.63 × 10 42 ) ( 23.8 MeV ) ⎜
⎝ 2⎠
⎝ 1 MeV ⎟⎠
(b)
The time this energy could possibly meet world requirements is
∆t =
E
P
=
⎛
⎞
2.53 × 10 31 J
1 yr
= ( 3.61 × 1016 s) ⎜
⎟
7
12
⎝ 3.16 × 10 s ⎠
100 ( 7.00 × 10 J s)
= 1.14 × 10 9 yr ~ 1 billion years
P45.19
(a)
Average KE per particle is
Therefore, vrms =
(b)
t=
3
1
kBT = m v 2 .
2
2
3kBT
=
m
0.1 m
x
~ 6
= ~ 10 −7 s
v 10 m s
3 (1.38 × 10 −23 J K ) ( 4.00 × 108 K )
2.014 (1.661 × 10 −27 kg )
= 2.22 × 10 6 m s
572
P45.20
Chapter 45
(a)
Including both ions and electrons, the number of particles in the plasma is N = 2nV where
n is the ion density and V is the volume of the container. Application of Equation 21.6 gives
the total energy as
E=
3
NkBT = 3nVkBT
2
⎡
⎛ 10 6 cm 3 ⎞ ⎤
−23
8
= 3 ( 2.0 × 1013 cm −3 ) ⎢ ( 50 m 3 ) ⎜
3
⎟⎠ ⎥ (1.38 × 10 J K ) ( 4.0 × 10 K )
⎝
1
m
⎣
⎦
E = 1.7 × 10 7 J
(b)
From Table 20.2, the heat of vaporization of water is Lv = 2.26 × 10 6 J kg . The mass of
water that could be boiled away is
m=
P45.21
(a)
1.7 × 10 7 J
E
=
= 7.3 kg
Lv 2.26 × 10 6 J kg
Lawson’s criterion for the D-T reaction is nτ ≥ 1014 s cm 3. For a confinement time of
τ = 1.00 s, this requires a minimum ion density of n = 1014 cm −3 .
(b)
At the ignition temperature of T = 4.5 × 10 7 K and the ion density found above, the plasma
pressure is
⎡
⎛ 10 6 cm 3 ⎞ ⎤
−23
7
P = 2nkBT = 2 ⎢ (1014 cm −3 ) ⎜
⎟⎠ ⎥ (1.38 × 10 J K ) ( 4.5 × 10 K )
3
⎝
1
m
⎣
⎦
= 1.24 × 10 5 J m 3
(c)
The required magnetic energy density is then
uB =
B2
≥ 10 P = 10 (1.24 × 10 5 J m 3 ) = 1.24 × 10 6 J m 3
2 µ0
B ≥ 2 ( 4π × 10 −7 N A 2 ) (1.24 × 10 6 J m 3 ) = 1.77 T This is a very strong field.
P45.22
The number of nuclei in 1.00 metric ton of trash is
N = 1 000 kg (1 000 g kg )
6.02 × 10 23 nuclei mol
= 1.08 × 10 28 nuclei
56.0 g mol
At an average charge of 26.0 e/nucleus,
Therefore t =
q = (1.08 × 10 28 ) ( 26.0 ) (1.60 × 10 −19 ) = 4.47 × 1010 C
q 4.47 × 1010
=
= 4.47 × 10 4 s = 12.4 h
I 1.00 × 10 6
Applications of Nuclear Physics
Section 45.5
P45.23
N0 =
573
Radiation Damage
mass present
5.00 kg
= 3.35 × 10 25 nuclei
=
mass of nucleus (89.907 7 u ) (1.66 × 10 −27 kg u )
λ=
ln 2
ln 2
=
= 2.38 × 10 −2 yr −1 = 4.53 × 10 −8 min −1
T1 2 29.1 yr
R0 = λ N 0 = ( 4.53 × 10 −8 min −1 ) ( 3.35 × 10 25 ) = 1.52 × 1018 counts min
R
10.0 counts min
= e− λ t =
= 6.59 × 10 −18
1.52 × 1018 counts min
R0
P45.24
and
λ t = − ln ( 6.59 × 10 −18 ) = 39.6
giving
t=
39.6
39.6
=
= 1.66 × 10 3 yr
λ
2.38 × 10 −2 yr −1
The source delivers 100 mrad of 2-MeV γ -rays/h at a 1.00-m distance.
(a)
For γ -rays, dose in rem = dose in rad.
Thus a person would have to stand there 10.0 hours to receive 1.00 rem from a
100-mrad/h source.
(b)
If the γ -radiation is emitted isotropically, the dosage rate falls off as
1
.
r2
Thus a dosage 10.0 mrad/h would be received at a distance r = 10.0 m = 3.16 m .
P45.25
(a)
The number of x-ray images made per year is
n = (8 x-ray d ) ( 5 d wk ) ( 50 wk yr ) = 2.0 × 10 3 x-ray yr
The average dose per photograph is
(b)
5.0 rem yr
= 2.5 × 10 −3 rem x-ray image
2.0 × 10 3 x-ray yr
The technician receives low-level background radiation at a rate of 0.13 rem yr . The dose
of 5.0 rem yr received as a result of the job is
5.0 rem yr
= 38 times the assumed background level . The technician’s occupational
0.13 rem yr
exposure is high compared to background radiation.
P45.26
(a)
I = I 0 e− µ x ,
so
x=
1 ⎛ I0 ⎞
ln
µ ⎝ I⎠
With µ = 1.59 cm −1 , the thickness when I =
(b)
When
I0
is
2
x=
1
ln ( 2 ) = 0.436 cm .
1.59 cm −1
I0
1
= 1.00 × 10 4 , x =
ln (1.00 × 10 4 ) = 5.79 cm .
I
1.59 cm −1
574
P45.27
Chapter 45
1 rad = 10 −2 J kg
∆t =
mc∆T
P
=
Q = mc∆T
P ∆ t = mc∆T
m ( 4 186 J kg ⋅°C) ( 50.0°C)
(10) (10−2
J kg ⋅ s) ( m )
= 2.09 × 10 6 s ≈ 24 days!
Note that power is the product of dose rate and mass.
P45.28
Q absorbed energy
10 −2 J kg
=
= (1 000 rad )
= 10.0 J kg
m
unit mass
1 rad
The rise in body temperature is calculated from Q = mc∆T where c = 4 186 J kg for water and
the human body
∆T =
Q
1
= 2.39 × 10 −3°C The temperature change is
= (10.0 J kg )
mc
4 186 J kg ⋅°C
negligible.
P45.29
If half of the 0.140-MeV gamma rays are absorbed by the patient, the total energy absorbed is
E=
( 0.140 MeV ) ⎡⎛ 1.00 × 10 −8 g ⎞ ⎛ 6.02 × 10 23 nuclei ⎞ ⎤
2
⎢⎜
⎟⎜
⎣⎝ 98.9 g mol ⎠ ⎝
1 mol
E = ( 4.26 × 1012 MeV ) (1.60 × 10 −13 J MeV ) = 0.682 J
Dose =
Thus, the dose received is
*P45.30 The decay constant is λ =
12
⎟⎠ ⎥ = 4.26 × 10 MeV
⎦
0.682 J ⎛ 1 rad ⎞
= 1.14 rad
60.0 kg ⎜⎝ 10 −2 J kg ⎟⎠
ln 2 ln 2
=
= 0.040 8 d . The number of nuclei remaining
T1 2 17 d
after 30 d is N = N 0 e− λT = N 0 e−0.040 8(30 ) = 0.294 N 0 . The number decayed is
N 0 − N = N 0 (1 − 0.294 ) = 0.706 N 0 . Then the energy release is
⎛ 1.6 × 10 −19 J ⎞
2.12 J = 0.706 N 0 ( 21.0 × 10 3 eV ) ⎜
⎟⎠
⎝
1 eV
N0 =
2.12 J
= 8.94 × 1014
2.37 × 10 −15 J
d ⎞
0.040 8
= 4.22 × 108 Bq
(8.94 × 1014 ) ⎛⎜⎝ 86 1400
d
s ⎟⎠
(a)
R0 = λ N 0 =
(b)
⎛ 1.66 × 10 −27 kg ⎞
original sample mass = m = N original mone atom = 8..94 × 1014 (103 u ) ⎜
⎟⎠
⎝
1u
= 1.53 × 10 −10 kg = 1.53 × 10 −7 g = 153 ng
Applications of Nuclear Physics
P45.31
The nuclei initially absorbed are
⎛ 6.02 × 10 23 nuclei mol ⎞
12
N 0 = (1.00 × 10 −9 g ) ⎜
⎟⎠ = 6.70 × 10
⎝
89.9 g moll
)
(
The number of decays in time t is ∆N = N 0 − N = N 0 (1 − e− λ t = N 0 1 − e−( ln 2)t T1 2
)
At the end of 1 year,
t
1.00 yr
=
= 0.034 4
T1 2 29.1 yr
and
∆N = N 0 − N = ( 6.70 × 1012 ) (1 − e−0.023 8 ) = 1.58 × 1011
The energy deposited is
E = (1.58 × 1011 ) (1.10 MeV ) (1.60 × 10 −13 J MeV ) = 0.027 7 J
Thus, the dose received is
⎛ 0.027 7 J ⎞
Dose = ⎜
= 3.96 × 10 −4 J kg = 0.039 6 rad
⎝ 70.0 kg ⎟⎠
Section 45.6
P45.32
575
Radiation Detectors
(a)
EI = 10.0 eV is the energy required to liberate an electron from a dynode. Let ni be the
number of electrons incident upon a dynode, each having gained energy e ( ∆V ) as it was
accelerated to this dynode. The number of electrons that will be freed from this dynode is
∆V
N i = ni e
:
EI
(1) e (100 V )
At the first dynode, ni = 1
and
N1 =
= 101 electrons
10.0 eV
(b)
For the second dynode, ni = N1 = 101 ,
so
N2 =
(101 )e (100 V )
= 10 2
10.0 eV
(10 2 )e (100 V )
= 10 3
10.0 eV
Observing the developing pattern, we see that the number of electrons incident on the
seventh and last dynode is n7 = N 6 = 10 6 .
At the third dynode, ni = N 2 = 10 2
(c)
and
N3 =
The number of electrons incident on the last dynode is n7 = 10 6. The total energy these
electrons deliver to that dynode is given by
E = ni e ( ∆V ) = 10 6 e ( 700 V − 600 V ) = 108 eV
576
Chapter 45
(a)
2
(1 2) ( 5.00 × 10 −12 F ) (1.00 × 103 V )
E (1 2 ) C ( ∆V )
=
=
= 3.12 × 10 7
Eβ
0.500 MeV
( 0.500 MeV ) (1.60 × 10 −13 J MeV )
(b)
N=
(a)
The average time between slams is 60 min 38 = 1.6 min. Sometimes, the actual interval
is nearly zero. Perhaps about equally as often, it is 2 × 1.6 min. Perhaps about half as often,
2
P45.33
P45.34
−12
3
Q C ( ∆V ) ( 5.00 × 10 F ) (1.00 × 10 V )
=
=
= 3.12 × 1010 electrons
e
e
1.60 × 10 −19 C
it is 4 × 1.6 min. Somewhere around 5 × 1.6 min = 8.0 min , the chances of randomness
producing so long a wait get slim, so such a long delay might likely be due to mischief.
(b)
The midpoints of the time intervals are separated by 5.00 minutes. We use R = R0 e− λ t .
Subtracting the background counts,
337 − 5 (15) = [ 372 − 5 (15)] e
2
)(5.00 min )
262 ⎞
= ln ( 0.882 ) = − 3.47 min T1 2 which yields T1 2 = 27.6 min
ln ⎛
⎝ 297 ⎠
or
(c)
(
− ln 2 T1
As in the random events in part (a), we imagine a ±5 count counting uncertainty. The
smallest likely value for the half-life is then given by
⎛ 262 − 5 ⎞
ln ⎜
⎟ = − 3.47 min T1 2 , or T1
⎝ 297 + 5 ⎠
( )
2 min
= 21.1 min
The largest credible value is found from
262 + 5 ⎞
= − 3.47 min T1 2 , yielding T1
ln ⎛
⎝ 297 − 5 ⎠
( )
Thus,
Section 45.7
P45.35
T1 2 = ⎛
⎝
2 max
= 38.8 min
38.8 + 21.1⎞ ⎛ 38.8 − 21.1⎞
min = ( 30 ± 9 ) min = 30 min ± 30%
±
⎠ ⎝
⎠
2
2
Uses of Radiation
The initial specific activity of
( R m )0 =
Fe in the steel,
20.0 µ Ci 100 µ Ci ⎛ 3.70 × 10 4 Bq ⎞
6
=
⎟⎠ = 3.70 × 10 Bq kg
0.200 kg
kg ⎜⎝
1 µ Ci
R ⎛ R ⎞ −λ t
−( 6.40 ×10 −4 h −1 )(1 000 h )
e = ( 3.70 × 10 6 Bq kg ) e
=
= 1.95 × 10 6 Bq kg
m ⎝ m⎠ 0
After 1 000 h,
The activity of the oil is
Therefore
59
min oil =
So that wear rate is
800
Roil = ⎛
Bq liter ⎞ (6.50 liters) = 86.7 Bq
⎝ 60.0
⎠
Roil
86.7 Bq
=
= 4.45 × 10 −5 kg
( R m ) 1.95 × 106 Bq kg
4.45 × 10 −5 kg
= 4.45 × 10 −8 kg h
1 000 h
Applications of Nuclear Physics
P45.36
(a)
577
10 4 MeV
= 9.62 × 10 3. Since only 50% of the photons are
1.04 MeV
detected, the number of 65 Cu nuclei decaying is twice this value, or 1.92 × 10 4. In two
The number of photons is
3
N 0 = 1.92 × 10 4 and
4
N 0 = 2.56 × 10 4. This is 1% of the 65 Cu, so the number of 65 Cu is 2.56 × 10 6 ~ 10 6 .
half-lives, three-fourths of the original nuclei decay, so
(b)
Natural copper is 69.17% 63 Cu and 30.83% 63 Cu . Thus, if the sample contains N Cu
copper atoms, the number of atoms of each isotope is
N 63 = 0.691 7 N Cu and N 65 = 0.308 3 N Cu
Therefore,
N 63 0.691 7
0.6917 ⎞
0.6917 ⎞
=
or N 63 = ⎛
N =⎛
2 .56 × 10 6 ) = 5.75 × 10 6.
⎝ 0.3083 ⎠ 65 ⎝ 0.3083 ⎠ (
N 65 0.308 3
mCu = ( 62 .93 u ) N 63 + ( 64.93 u ) N 65 :
The total mass of copper present is then
6
6
mCu = ⎡⎣( 62 .93) ( 5.75 × 10 ) + ( 64.93) ( 2 .56 × 10 )⎤⎦ u (1.66 × 10 −24 g u )
= 8.77 × 10 −16 g ~ 10 −15 g
P45.37
(a)
Starting with N = 0 radioactive atoms at t = 0, the rate of increase is (production – decay)
dN
= R−λN
dt
so
dN = ( R − λ N ) dt
The variables are separable.
N
dN
t
∫ R − λ N = ∫ dt :
0
(b)
−
0
so
⎛ R − λN ⎞
ln ⎜
= −λ t
⎝ R ⎟⎠
Therefore
1−
λ
N = e− λ t
R
and
1 ⎛ R−λ N⎞
ln ⎜
⎟ =t
λ ⎝ R ⎠
⎛ R − λN ⎞
−λ t
⎟ =e
⎜⎝
R ⎠
N=
R
(1 − e− λ t )
λ
The maximum number of radioactive nuclei would be
R
.
λ
578
Chapter 45
*P45.38 (a)
(b)
With I ( x ) =
1
I 0 , I ( x ) = I 0 e− µ x becomes
2
1
I 0 = I 0 e−0.72 x mm
2
2 = e+0.72 x mm
ln 2 = 0.72 x mm
x=
ln 2 mm
= 0.963 mm
0.72
I ( 0.8 mm ) = I 0 e−0.72( 0.8) = 0.562 I 0
I ( 0.7 mm ) = I 0 e−0.72( 0.7) = 0.604 I 0
fractional change =
0.604 I 0 − 0.562 I 0
= 0.074 7 = 7.47%
0.562 I 0
Additional Problems
P45.39
(a)
At 6 × 108 K, the average kinetic energy of a carbon atom is
3
kBT = (1.5) (8.62 × 10 −5 eV K ) ( 6 × 108 K ) = 8 × 10 4 eV
2
Note that 6 × 108 K is about 6 2 = 36 times larger than 1.5 × 10 7 K, the core temperature
of the Sun. This factor corresponds to the higher potential-energy barrier to carbon fusion
compared to hydrogen fusion. It could be misleading to compare it to the temperature
~10 8 K required for fusion in a low-density plasma in a fusion reactor.
(b)
The energy released is
E = ⎡⎣ 2 m ( C12 ) − m ( Ne 20 ) − m ( He 4 ) ⎤⎦ c 2
E = ( 24.000 000 − 19.992 440 − 4.002 603) ( 931.5) MeV = 4.62 MeV
In the second reaction,
E = ⎡⎣ 2 m ( C12 ) − m ( Mg24 ) ⎤⎦ ( 931.5) MeV u
E = ( 24.000 000 − 23.985 042 ) ( 931.5) MeV = 13.9 MeV
(c)
The energy released is the energy of reaction of the number of carbon nuclei in a 2.00-kg
sample, which corresponds to
⎛ 6.02 × 10 23 atoms mol ⎞ ⎛ 4.62 MeV fusion event ⎞ ⎛
⎞
1 kWh
∆E = ( 2.00 × 10 3 g ) ⎜
⎟⎜
⎟⎜
⎟
19
12.0 g mol
⎝
⎠ ⎝ 2 nuclei fusion eveent ⎠ ⎝ 2.25 × 10 MeV ⎠
(1.00 × 10 ) ( 4.62) kWh =
2 ( 2.25 × 10 )
26
∆E =
19
1.03 × 10 7 kWh
Applications of Nuclear Physics
P45.40
579
To conserve momentum, the two fragments must move in opposite directions with speeds v1 and
v2 such that
m1v1 = m2 v2
⎛m ⎞
v2 = ⎜ 1 ⎟ v1
⎝ m2 ⎠
or
The kinetic energies after the break-up are then
2
K1 = 1 m1v12
2
⎛m ⎞
⎛m ⎞
K 2 = 12 m2 v22 = 12 m2 ⎜ 1 ⎟ v12 = ⎜ 1 ⎟ K1
⎝ m2 ⎠
⎝ m2 ⎠
and
The fraction of the total kinetic energy carried off by m1 is
K1
K1
m2
=
=
K1 + K 2 K1 + ( m1 m2 ) K1
m1 + m2
and the fraction carried off by m2 is 1 −
P45.41
(a)
m2
m1
=
.
m1 + m2
m1 + m2
Q = 236.045 562 u c 2 − 86.920 711u c 2 − 148.934 370 u c 2 = 0.190 481u c 2 = 177 MeV
Immediately after fission, this Q-value is the total kinetic energy of the fission products.
(b)
⎛ mLa ⎞
K Br = ⎜
⎟ Q , from Problem 45.40
⎝ mBr + mLa ⎠
⎛ 149 u ⎞
=⎜
⎟ (177.4 MeV) = 112 MeV
⎝ 87 u + 149 u ⎠
K La = Q − K Br = 177.4 MeV − 112.0 MeV = 65.4 MeV
(c)
P45.42
vBr =
2 (112 × 10 6 eV) (1.6 × 10 −19 J eV)
2 K Br
= 1.58 × 10 7 m s
=
mBr
(87 u) (1.66 × 10−27 kg u)
vLa =
2 ( 65.4 × 10 6 eV) (1.6 × 10 −19 J eV)
2 K La
= 9.20 × 10 6 m s
=
mLa
(149 u) (1.66 × 10−27 kg u)
For a typical 235 U fission, Q = 208 MeV and the initial mass is 235 u. Thus, the fractional
energy loss is
Q
208 MeV
= 9.50 × 10 − 4 = 0.095 0%
=
mc 2 ( 235 u ) ( 931.5 MeV u )
For the D-T fusion reaction,
Q = 17.6 MeV
The initial mass is
m = ( 2.014 u ) + ( 3.016 u ) = 5.03 u
The fractional loss in this reaction is
Q
17.6 MeV
=
= 3.75 × 10 − 3 = 0.375%
mc 2 ( 5.03 u )( 931.5 MeV u )
0.375%
= 3.95 or the fractional loss in D-T fusion is about 4 times that in 235 U fission .
0.0950%
580
P45.43
Chapter 45
The decay constant is λ =
ln 2
ln 2
=
= 1.78 × 10 −9 s−1.
7
T1 2 (12.3 yr ) ( 3.16 × 10 s yr )
The tritium in the plasma decays at a rate of
⎡⎛ 2.00 × 1014 ⎞ ⎛ 10 6 cm 3 ⎞
3 ⎤
R = λ N = (1.78 × 10 − 9 s−1 ) ⎢⎜
⎟⎜
⎟ ( 50.0 m )⎥
3
3
1
m
cm
⎠⎝
⎠
⎣⎝
⎦
⎛
1 Ci
⎞
R = 1.78 × 1013 Bq = (1.78 × 1013 Bq ) ⎜
= 482 Ci
10
⎝ 3.70 × 10 Bq ⎟⎠
The fission inventory is
4 × 1010 Ci
~ 10 8 times greater than this amount.
482 Ci
5 × 10 6 Ci ⎛ 1 km ⎞
= 5 × 10 −4 Ci/m 2.
10 4 km 2 ⎝ 10 3 m ⎠
2
*P45.44 The original activity per area is
The decay constant is l = ln 2/27.7 yr.
The decay law N = N0 e–lt becomes the law of decrease of activity R = R0 e–lt. If the material is
not transported, it describes the time evolution of activity per area R/A = R0/A e–lt. Solving gives
⎛ R /A ⎞
ln 2
t
λt = ln ⎜ 0 ⎟ =
/
27
.7 yr
R
A
⎝
⎠
27.7 yr ⎛ R0 /A ⎞ 27.7 yr ⎛ 5 × 10 −4 Ci/m 2 ⎞
= 221 yr
t=
ln ⎜
ln ⎜
⎟=
−6
2 ⎟
ln 2
ln 2
⎝ 2 × 10 Ci/m ⎠
⎝ R/A ⎠
e λt =
P45.45
R0 /A
R/A
The complete fissioning of 1.00 gram of U 235 releases
Q=
(1.00 g)
(6.02 × 10 23 atoms mol ) ( 200 MeV fission ) (1.60 × 10 −13 J MeV )
235 grams mol
= 8.20 × 1010 J
If all this energy could be utilized to convert m kilograms of 20.0°C water to 400°C steam (see
Chapter 20 of text for values),
then
Q = mcw ∆T + mLv + mcs ∆T
Q = m ⎡⎣( 4186 J kg °C) ( 80.0 °C) + 2.26 × 10 6 J kg + ( 2010 J kg °C) ( 300 °C)⎤⎦
Therefore
m=
8.20 × 1010 J
= 2.56 × 10 4 kg
3.20 × 10 6 J kg
Applications of Nuclear Physics
P45.46
581
When mass m of 235 U undergoes complete fission, releasing 200 MeV per fission event, the total
energy released is:
⎛
⎞
m
Q=⎜
⎟ N A ( 200 MeV) where N A is Avogadro’s number.
⎝ 235 g mol ⎠
If all this energy could be utilized to convert a mass mw of liquid water at Tc into steam at Th,
then Q = m w ⎡⎣c w (100°C − Tc ) + L v + c s (Th − 100°C)⎤⎦
where c w is the specific heat of liquid water, Lv is the latent heat of vaporization, and c s is the
specific heat of steam. Solving for the mass of water converted gives
mw =
=
*P45.47 (a)
Q
⎡⎣c w (100°C − Tc ) + L v + c s (Th − 100°C)⎤⎦
(235
mN A ( 200 MeV)
g mol) ⎡⎣cw (100°C − Tc ) + Lv + cs (Th − 100°C)⎤⎦
.
We have 0.9928 kg of 238U, comprising
N = 0.9928 kg (6.02 × 1023 atoms/mol)/(0.238 kg/mol) = 2.51 × 1024 nuclei, with activity
R = λ N = (ln 2/T1/2)N = 2.51 × 1024 (ln 2/4.47 × 109 yr)(1 yr/3.16 × 107 s)(1 Ci/3.7 × 1010/s)
= 333 µCi
We have 0.0072 kg of 235U, comprising
N = 0.0072 kg (6.02 × 1023 atoms/mol)/(0.235 kg/mol) = 1.84 × 1022 nuclei, with activity
R = λN = (ln 2/T1/2)N = 1.84 × 1022 (ln 2/7.04 × 108 yr)(1 yr/3.16 × 107 s)(1 Ci/3.7 × 1010/s)
= 15.5 µCi
We have 0.00005 kg of 234U, comprising
N = 5 × 10–5 kg (6.02 × 1023 atoms/mol)/(0.23404 kg/mol) = 1.29 × 1020 nuclei, with activity
R = λ N = (ln 2/T1/2)N = 1.29 × 1020 (ln 2/2.44 × 105 yr)(1 yr/3.16 × 107 s)(1 Ci/3.7 × 1010/s)
= 312 µCi
P45.48
(b)
The total activity is (333 + 15.5 + 312) µCi = 661 µCi, so the fractional contributions are
respectively 333/661 = 50.4% 15.5/661 = 2.35% and 312/661 = 47.3%
(c)
It is potentially dangerous, notably if the material is inhaled as a powder. With precautions
to minimize human contact, microcurie sources are routinely used in laboratories.
(a)
∆V = 4 π r 2 ∆ r = 4 π (14.0 × 10 3 m ) ( 0.05 m ) = 1.23 × 10 8 m 3 ~10 8 m 3
(b)
The force on the next layer is determined by atmospheric pressure.
2
W = P∆V = (1.013 × 10 5 N/m 2 ) (1.23 × 10 8 m 3 ) = 1.25 × 1013 J ~1013 J
(c)
(d)
1
(yield) , so yield = 1.25 × 1014 J ~1014 J
10
1.25 × 1014 J
= 2.97 × 10 4 ton TNT ~ 10 4 ton TNT
4.2 × 10 9 J ton TNT
1.25 × 1013 J =
or ~ 10 kilotons
582
P45.49
Chapter 45
(a)
The number of molecules in 1.00 liter of water (mass = 1 000 g) is
⎛ 1.00 × 10 3 g ⎞
23
25
N= ⎜
⎟ ( 6.02 × 10 moleculles mol) = 3.34 × 10 molecules.
18.0
g
mol
⎝
⎠
The number of deuterium nuclei contained in these molecules is
⎛ 1 deuteron ⎞
22
N ′ = ( 3.34 × 10 25 molecules) ⎜
⎟ = 1.01 × 10 deuterons.
⎝ 3300 moleccules ⎠
Since 2 deuterons are consumed per fusion event, the number of events possible is
N′
= 5.07 × 10 21 reactions, and the energy released is
2
Efusion = ( 5.07 × 10 21 reactions) ( 3.27 MeV reaction ) = 1.66 × 10 22 MeV
Efusion = (1.66 × 10 22 MeV) (1.60 × 10 −13 J MeV) = 2.665 × 10 9 J .
(b)
In comparison to burning 1.00 liter of gasoline, the energy from the fusion of deuterium is
Efusion 2.65 × 10 9 J
=
= 78.0 times larger .
Egasoline 3.40 × 10 7 J
P45.50
Momentum conservation:
0 = mLi v Li + mα v α
Thus,
K Li =
or
mLi vLi = mα vα
2
2
(m v ) ⎛ m2 ⎞
1
1 ( mLi vLi )
mLi vLi2 =
= α α = ⎜ α ⎟ vα2
2
2 mLi
2 mLi
⎝ 2mLi ⎠
⎛ ( 4.002 6 u ) 2 ⎞
2
2
⎟ ⎛⎜ 9.25 × 10 6 m s ⎞⎟ = (1.14 u ) ⎛⎜ 9.25 × 10 6 m s ⎞⎟
K Li = ⎜⎜
⎝
⎠
⎝
⎠
⎟
2
7.016
0
u
) ⎠
⎝ (
K Li = 1.14 (1.66 × 10 − 27 kg ) ⎜⎝ 9.25 × 10 6 m s ⎟⎠ = 1.62 × 10 −13 J = 1.01 MeV .
⎞2
⎛
P45.51
(a)
The thermal power transferred to the water is
Pw = 0.970 ( waste heat )
Pw = 0.970 ( 3 065 − 1 000 ) MW = 2 .00 × 10 9 J s
rw is the mass of water heated per hour:
rw =
The volume used per hour is
(b)
The
235
Pw
c ( ∆T )
=
( 2 .00 × 10
( 4186
9
J s ) ( 3600 s h )
J kg ⋅ °C
C ) ( 3.50 °C )
4.91 × 108 kg h
= 4.91 × 10 5 m 3 h
1.00 × 10 3 kg m 3
U fuel is consumed at a rate
⎛ 3 065 × 10 6 J s ⎞ ⎛ 1 kg ⎞ ⎛ 3 600 s ⎞
rf = ⎜
= 0.141 kg h
⎝ 7.80 × 1010 J g ⎟⎠ ⎜⎝ 1 000 g ⎟⎠ ⎝ 1 h ⎠
= 4.91 × 108 kg h
Applications of Nuclear Physics
P45.52
210
The number of nuclei in 0.155 kg of
583
Po is
⎛
155 g
⎞
N0 = ⎜
(6.02 × 10 23 nuclei mol ) = 4.44 × 10 23 nuclei
⎝ 209.98 g mol ⎟⎠
The half-life of
λ=
210
Po is 138.38 days, so the decay constant is given by
ln 2
ln 2
=
= 5.80 × 10 −8 s −1
T1 2 (138.38 d ) (8.64 × 10 4 s d )
The initial activity is
R0 = λ N 0 = ( 5.80 × 10 −8 s −1 ) ( 4.44 × 10 23 nuclei ) = 2.58 × 1016 Bq
The energy released in each
210
84
4
Po → 206
82 Pb + 2 He reaction is
Q = ⎡⎣ M 210 Po − M 206 Pb − M 4 He ⎤⎦ c 2 :
84
82
2
Q = [ 209.982 857 − 205.974 449 − 4.002 603] u ( 931.5 MeV
V u ) = 5.41 MeV
Thus, assuming a conversion efficiency of 1.00%, the initial power output of the battery is
P = ( 0.010 0 ) R0Q = ( 0.010 0 ) ( 2.58 × 1016 decays s ) ( 5.41 MeV decay ) (1.60 × 10 −13 J MeV )
= 223 W
P45.53
(a)
(b)
⎛m⎞
m
, so = ⎜ ⎟
ρ
⎝ρ⎠
1 3
13
⎛
⎞
70.0 kg
= 0.155 m
=⎜
3
3⎟
⎝18.7 × 10 kg m ⎠
Add 92 electrons to both sides of the given nuclear reaction. Then it becomes
V = 3 =
238
92
U atom → 8 42 He atom +
206
82
Pb atom + Qnet /c 2
Qnet = ⎡ M 238 U − 8 M 4 He − M 206 Pb ⎤ c 2
2
82
⎣ 92
⎦
= ⎡⎣ 238.050 783 − 8 ( 4.002 603) − 205.974 449 ⎤⎦ u ( 931.5 MeV u )
Qnet = 51.7 MeV
(c)
If there is a single step of decay, the number of decays per time is the decay rate R and the
energy released in each decay is Q. Then the energy released per time is P = QR . If
there is a series of decays in steady state, the equation is still true, with Q representing the
net decay energy.
(d)
The decay rate for all steps in the radioactive series in steady state is set by the parent
uranium:
⎛ 7.00 × 10 4 g ⎞
26
23
N=⎜
⎟ (6.02 × 10 nuclei mool ) = 1.77 × 10 nuclei
⎝ 238 g mol ⎠
ln 2
ln 2
1
λ=
=
= 1.55 × 10 −10
T1 2 4.47 × 10 9 yr
yr
⎛
1⎞
R = λ N = ⎜ 1.55 × 10 − 10
(1.77 × 10 26 nuclei ) = 2 .75 × 1016 decays yr
⎝
yr ⎟⎠
so
⎛
P = QR = ( 51.7 MeV) ⎜ 2 .75 × 1016
⎝
1⎞
−13
5
⎟ (1.60 × 10 J MeV) = 2.27 × 10 J yr
yr ⎠
continued on next page
584
Chapter 45
dose in rem = dose in rad × RBE
(e)
5.00 rem yr = ( dose in rad yr ) 1.10, giving ( dose in rad yr ) = 4.55 rad yr
⎛ 10 −2 J kg ⎞
The allowed whole-body dose is then ( 70.0 kg )( 4.55 rad yr ) ⎜
⎟ = 3.18 J yr .
⎝ 1 rad ⎠
P45.54
ET ≡ E ( thermal) =
3
kBT = 0.039 eV
2
n
1
1
ET = ⎛⎜ ⎞⎟ E where n ≡ number of collisions, and 0.039 = ⎛ ⎞
⎝ 2⎠
⎝2⎠
n
( 2.0 × 10 )
6
Therefore n = 25.6 = 26 collisions
P45.55
Conservation of linear momentum and energy can be applied to find the kinetic energy of the
neutron. We first suppose the particles are moving nonrelativistically.
The momentum of the alpha particle and that of the neutron must add to zero, so their velocities
must be in opposite directions with magnitudes related by
mn v n + mα vα = 0
or
(1.008 7 u ) vn = ( 4.002 6 u ) vα
At the same time, their kinetic energies must add to 17.6 MeV
E=
1
1
1
1
mn vn2 + mα vα2 = (1.008 7 u ) vn2 + ( 4.002 6 ) vα2 = 17.6 MeV
2
2
2
2
⎞
⎛
1u
Substitute vα = 0.252 0 vn: E = ( 0.504 35 u ) vn2 + ( 0.127 10 u ) vn2 = 17.6 MeV ⎜
2 ⎟
⎝ 931.494 MeV c ⎠
vn =
0.018 9c 2
= 0.173c = 5.19 × 10 7 m s
0.631 45
Since this speed is not too much greater than 0.1c, we can get a reasonable estimate of the kinetic
energy of the neutron from the classical equation,
K=
2
1
1
2 ⎛ 931.494 MeV c ⎞
m v 2 = (1.008 7 u ) ( 0.173c ) ⎜
⎟⎠ = 14.1 MeV
⎝
2
2
u
For a more accurate calculation of the kinetic energy, we should use relativistic expressions.
Conservation of momentum gives
γ n mn v n + γ α mα vα = 0
1.008 7
yielding
vα2
vn2
=
2
2
c
15.746c − 14.746 vn2
Then
(γ n − 1) mn c2 + (γ α − 1) mα c2 = 17.6 MeV
and
2
vn = 0.171c implying that (γ n − 1) mn c = 14.0 MeV
vn
1− v c
2
n
2
= 4.002 6
vα
1 − vα2 c 2
Applications of Nuclear Physics
P45.56
From the table of isotopic masses in Chapter 44, the half-life of
constant is
λ=
32
P is 14.26 d. Thus, the decay
ln 2
ln 2
=
= 0.048 6 d −1 = 5.63 × 10 −7 s −1
T1 2 14.26 d
N0 =
R0 5.22 × 10 6 decay s
=
= 9.28 × 1012 nuclei
λ
5.63 × 10 −7 s −1
At t = 10.0 days, the number remaining is
− 0.048 6 d −1 )(10 . 0 d )
N = N 0 e− λ t = ( 9.28 × 1012 nuclei ) e (
= 5.71 × 1012 nuclei
so the number of decays has been N 0 − N = 3.57 × 1012 and the energy released is
E = ( 3.57 × 1012 ) ( 700 keV ) (1.60 × 10 −16 J keV ) = 0.400 J
If this energy is absorbed by 100 g of tissue, the absorbed dose is
⎛ 0.400 J ⎞ ⎛ 1 rad ⎞
Dose = ⎜
= 400 rad
⎝ 0.100 kg ⎟⎠ ⎜⎝ 10 −2 J kg ⎟⎠
P45.57
(a)
The number of Pu nuclei in 1.00 kg =
6.02 × 10 23 nuclei mol
(1 000 g ) .
239.05 g mol
The total energy = ( 25.2 × 10 23 nuclei ) ( 200 MeV ) = 5.04 × 10 26 MeV
E = (5.04 × 10 26 MeV ) ( 4.44 × 10 −20 kWh MeV ) = 2.24 × 10 7 kWh
or
(b)
22 million kWh
E = ∆m c 2 = (3.016 049 u + 2.014 102 u − 4.002 603 u − 1.008 665 u )( 931.5 MeV u )
E = 17.6 MeV for each D-T fusion
(c)
En = ( Total number of D nuclei )(17.6 ) ( 4.44 × 10 −20 )
1 000 ⎞
⎟ (17.6 ) ( 4.44 × 10 −20 ) = 2.34 × 108 kWh
En = (6.02 × 10 23 ) ⎛⎜
⎝ 2.014 ⎠
(d)
En = the number of C atoms in 1.00 kg × 4.20 eV
⎛ 6.02 × 10 26 ⎞
−6
−20
En = ⎜
⎟ ( 4.20 × 10 MeV ) ( 4.44 × 10 ) = 9.36 kWh
12
⎝
⎠
(e)
585
Coal is cheap at this moment in human history. We hope that safety and waste disposal
problems can be solved so that nuclear energy can be affordable before scarcity drives
up the price of fossil fuels. Burning coal in the open puts carbon dioxide into the
atmosphere, worsening global warming. Plutonium is a very dangerous material to have
sitting around.
586
P45.58
Chapter 45
Add two electrons to both sides of the given reaction.
4 11 H atom → 42 He atom + Q / c 2
Then
where
Q = ( ∆m ) c 2 = ⎡⎣ 4 (1.007 825) − 4.002 603 ⎤⎦ u ( 931.5 MeV
V u ) = 26.7 MeV
or
Q = ( 26.7 MeV ) (1.60 × 10 −13 J MeV ) = 4.28 × 10 −12 J
The proton fusion rate is then
rate =
P45.59
(a)
power output
3.85 × 10 26 J s
=
= 3.60 × 10 38 protons s
energy per proton ( 4.28 × 10 −12 J ) ( 4 protons )
QI = [ M A + M B − M C − M E ] c 2 ,
and
QII = [ M C + M D − M F − M G ] c 2
Qnet = QI + QII = [ M A + M B − M C − M E + M C + M D − M F − M G ] c 2
Qnet = QI + QII = [ M A + M B + M D − M E − M F − M G ] c 2
Thus, reactions may be added. Any product like C used in a subsequent reaction does not
contribute to the energy balance.
(b)
Adding all five reactions gives
1
1
or
H + 11 H+ −01 e + 11 H+ 11 H+ −01 e → 42 He + 2ν + Qnet /c 2
4 11 H + 2
0
−1
e → 42 He + 2ν + Qnet /c 2
Adding two electrons to each side gives
4 11 H atom → 42 He atom + Qnet /c 2
Thus Qnet = ⎡⎣4 M 1 H − M 4 He ⎤⎦ c 2 = ⎡⎣4 (1.007 825 ) − 4.002 603⎤⎦ u ( 931.5 MeV u )
1
2
= 26.7 MeV
P45.60
(a)
⎡ 4π ⎛1.50 × 10 − 2 cm ⎞ 3 ⎤
−7
The mass of the pellet is m = ρV = ( 0.200 g cm ) ⎢
⎜
⎟ ⎥ = 3.53 × 10 g
2
⎠ ⎥⎦
⎢⎣ 3 ⎝
3
The pellet consists of equal numbers of 2 H and 3 H atoms, so the average molar mass is
2.50 and the total number of atoms is
⎛ 3.53 × 10 − 7 g ⎞
23
16
N=⎜
⎟ (6.02 × 10 atoms mol ) = 8.51 × 10 atoms
⎝ 2 .50 g mol ⎠
When the pellet is vaporized, the plasma will consist of 2N particles (N nuclei and
N electrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. The
temperature of the plasma is found from E = ( 2 N ) 23 kBT as
(
T=
(b)
)
E
2 .00 × 10 3 J
=
= 5.68 × 108 K
3 NkB 3 (8.51 × 1016 ) (1.38 × 10 − 23 J K )
Each fusion event uses 2 nuclei, so
N
events will occur. The energy released will be
2
⎛ 8.51 × 1016 ⎞
N
−13
5
E =⎛ ⎞Q =⎜
⎟⎠ (17.59 MeV ) (1.60 × 10 J MeV ) = 1.20 × 10 J = 120 kJ
⎝ 2⎠
⎝
2
Applications of Nuclear Physics
P45.61
587
(a)
The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome
k ( e )( 2 e )
the Coulomb-repulsion barrier to 11 H + 23 He → 42 He + e+ + ν , estimated as e
. The
r
7
k ( e )( 7e )
Coulomb barrier to Bethe’s fifth and eight reactions is like e
times,
, larger by
2
r
7
so the required temperature can be estimated as (15 × 10 6 K ) ≈ 5 × 10 7 K .
2
(b)
For
12
C + 1 H → 13 N + Q,
Q1 = (12.000 000 + 1.007 825 − 13.005 739 ) ( 931.5 MeV ) = 1.94 MeV
For the second step, add seven electrons to both sides to have:
13
N atom → 13 C atom + e+ + e− + Q
Q2 = ⎡⎣ 13.005 739 − 13.003 355 − 2 ( 0.000 549 )⎤⎦ ( 931.5 MeV ) = 1.20 MeV
Q3 = Q7 = 2 ( 0.000 549 ) ( 931.5 MeV ) = 1.02 MeV
Q4 = [13.003 355 + 1.007 825 − 14.003 074 ] ( 931.5 MeV ) = 7.55 MeV
Q5 = [14.003 074 + 1.007 825 − 15.003 065] ( 931.5 MeV ) = 7.30 MeV
Q6 = ⎡⎣ 15.003 065 − 15.000 109 − 2 ( 0.000 549 )⎤⎦ ( 931.5 MeV ) = 1.73 MeV
Q8 = [15.000 109 + 1.007 825 − 12 − 4.002 603] ( 931.5 MeV ) = 4.97 MeV
The sum is 26.7 MeV , the same as for the proton-proton cycle.
(c)
P45.62
Not quite all of the energy released appears as internal energy in the star. When a neutrino
is created, it will likely fly directly out of the star without interacting with any other
particle.
(a)
I 2 I 0 e − µ2 x
=
= e−(µ2 − µ1 )x
I1 I 0 e− µ1x
(b)
I 50
= e−(5.40− 41.0 )(0.100 ) = e3.56 = 35.2
I100
(c)
I 50
= e−(5.40− 41.0 )(1.00 ) = e35.6 = 2.89 × 1015
I100
Thus, a 1.00-cm-thick aluminum plate has essentially removed the long-wavelength x-rays
from the beam.
588
P45.63
Chapter 45
(a)
The number of fissions occurring in the zeroth, first, second, . . . nth generation is
N 0 , N 0 K , N 0 K 2 , . . ., N 0 K n
The total number of fissions that have occurred up to and including the nth generation is
N = N 0 + N 0 K + N 0 K 2 + + N 0 K n = N 0 (1 + K + K 2 + + K n )
Note that the factoring of the difference of two squares, a 2 − 1 = ( a + 1) ( a − 1) , can be
generalized to a difference of two quantities to any power,
a 3 − 1 = ( a 2 + a + 1) ( a − 1)
a n+1 − 1 = ( a n + a n−1 + + a 2 + a + 1) ( a − 1)
K n + K n −1 ++ K 2 + K + 1 =
Thus
N = N0
and
(b)
K n+1 − 1
K −1
K n+1 − 1
K −1
The number of U-235 nuclei is
1 atom ⎞ ⎛
1u
⎞
N = 5.50 kg ⎛
= 1.41 × 10 25 nuclei
⎝ 235 u ⎠ ⎜⎝ 1.66 × 10 −27 kg ⎟⎠
We solve the equation from part (a) for n, the number of generations:
N
( K − 1) = K n +1 − 1
N0
N
( K − 1) + 1 = K n ( K )
N0
⎛ N ( K − 1) ⎞
⎛ N ( K − 1) N 0 + 1 ⎞
n ln K = ln ⎜
+ 1⎟ − ln K
⎟ = ln ⎜
K
⎝
⎠
⎝ N0
⎠
n=
ln (1.41 × 10 25 ( 0.1) 10 20 + 1)
ln ( N ( K − 1) N 0 + 1)
−1=
− 1 = 99.2
ln 1.1
ln K
Therefore time must be allotted for 100 generations:
∆tb = 100 (10 × 10 −9 s ) = 1.00 × 10 −6 s
(c)
v=
(d)
V=
150 × 10 9 N m 2
B
=
= 2.83 × 10 3 m s
ρ
18.7 × 10 3 kg m 3
4 3 m
πr =
3
ρ
⎛ 3m ⎞
r =⎜
⎟
⎝ 4 πρ ⎠
13
∆t d =
⎞
⎛
3 ( 5.5 kg )
⎟
=⎜
3
3
⎜ 4 π 18.7 × 10 kg m ⎟
)⎠
⎝ (
13
r 4.13 × 10 −2 m
=
= 1.46 × 10 −5 s
v 2.83 × 10 3 m s
continued on next page
= 4.13 × 10 −2 m
Applications of Nuclear Physics
(e)
14.6 µs is greater than 1 µs , so the entire bomb can fission. The destructive energy
released is
⎛ 200 × 10 6 eV ⎞ ⎛ 1.6 × 10 −19
1.41 × 10 25 nuclei ⎜
⎟⎜
⎝ fissioning nucleus ⎠ ⎝ 1 eV
J⎞
14
⎟ = 4.51 × 10 J
⎠
⎛ 1 ton TNT ⎞
= 4.51 × 1014 J ⎜
⎟
⎝ 4.2 × 10 9 J ⎠
= 1.07 × 10 5 ton TNT
= 107 kilotons of TNT
20
What if? If the bomb did not have an “initiator” to inject 10 neutrons at the moment
when the critical mass is assembled, the number of generations would be
n=
ln (1.41 × 10 25 ( 0.1) 1 + 1)
ln 1.1
− 1 = 582 requiring 583 (10 × 10 −9 s ) = 5.83 µs
This time is not very short compared with 14.6 µs, so this bomb would likely release
much less energy.
ANSWERS TO EVEN PROBLEMS
P45.2
184 MeV
P45.4
(a) 173 MeV
P45.6
(a) 8.68 MeV (b) The proton and the boron nucleus have positive charges. The
colliding particles must have enough kinetic energy to approach very closely in spite
of their electric repulsion.
P45.8
2.63 kg d
P45.10
(a) 4.84V −1 3 (b) 6V −1 3 (c) 6.30V −1 3
parallelepiped maximum.
P45.12
6.25 × 1019 Bq
P45.14
25.6 MeV
P45.16
(a) (144 keV) Z1 Z2 (b) The energy is proportional to each atomic number. (c) Take Z1 =1
and Z2 = 59 or vice versa. This choice minimizes the product Z1 Z2. (d) 144 keV for both,
according to this model
P45.18
(a) 2.53 × 10 31 J
P45.20
(a) 1.7 × 107 J
P45.22
12.4 h
P45.24
(a) 10.0 h (b) 3.16 m
P45.26
(a) 0.436 cm
(b) 0.078 8%
(b) 1.14 × 10 9 yr
(b) 7.3 kg
(b) 5.79 cm
(d) The sphere has minimum loss and the
589
590
Chapter 45
P45.28
2.39 × 10 −3 ° C
P45.30
(a) 422 MBq
P45.32
(a) 10 (b) 10 6
P45.34
(a) about 8 min (b) 27.6 min (c) 30 min ± 30%
P45.36
(a) ~10 6
P45.38
(a) 0.963 mm
P45.40
See the solution.
P45.42
The fractional loss in D - T is about 4 times that in 235 U fission.
P45.44
221 yr
P45.46
(b) 153 ng
(c) 108 eV
(b) ~10 −15 g
(b) It increases by 7.47%.
mN A ( 200 MeV)
(235
g mol) ⎡⎣cw (100°C − Tc ) + Lv + cs (Th − 100°C)⎤⎦
P45.48
(a) ~10 8 m 3 (b) ~1013 J
P45.50
1.01 MeV
P45.52
223 W
P45.54
26 collisions
P45.56
400 rad
P45.58
3.60 × 10 38 protons s
P45.60
(a) 5.68 × 108 K
P45.62
(a) See the solution.
(c) ~1014 J
(d) ~10 kilotons
(b) 120 kJ
(b) 35.2
(c) 2.89 × 1015
46
Particle Physics and Cosmology
CHAPTER OUTLINE
The Fundamental Forces in
Nature
46.2 Positrons and Other Antiparticles
46.3 Mesons and the Beginning of
Particle Physics
46.4 Classification of Particles
46.5 Conservation Laws
46.6 Strange Particles and
Strangeness
46.7 Finding Patterns in the Particles
46.8 Quarks
46.9 Multicolored Quarks
46.10 The Standard Model
46.11 The Cosmic Connection
46.12 Problems and Perspectives
ANSWERS TO QUESTIONS
46.1
Q46.1
*Q46.2 Answer (b). The electron and positron together have
very little momentum. A 1.02-MeV photon has a definite
chunk of momentum. Production of a single gamma ray
could not satisfy the law of conservation of momentum,
which must hold true in this—and every—interaction.
Q46.3
Q46.4
Strong Force—Mediated by gluons.
Electromagnetic Force—Mediated by photons.
Weak Force—Mediated by W+, W− , and Z 0 bosons.
Gravitational Force—Mediated by gravitons.
Hadrons are massive particles with structure and size.
There are two classes of hadrons: mesons and baryons.
Hadrons are composed of quarks. Hadrons interact via
the strong force. Leptons are light particles with no
structure or size. It is believed that leptons are fundamental particles. Leptons interact via the weak force.
1
3
or
composed of three quarks. Mesons are light
2
2
hadrons with spin 0 or 1 composed of a quark and an antiquark.
Baryons are heavy hadrons with spin
*Q46.5 Answer (c). The muon has much more rest energy than the electron and the neutrinos together.
The missing rest energy goes into kinetic energy.
*Q46.6 Answer (b). The z component of its angular momentum must be 3/2, 1/2, –1/2, or –3/2, in
units of .
Q46.7
The baryon number of a proton or neutron is one. Since baryon number is conserved, the baryon
number of the kaon must be zero.
Q46.8
Decays by the weak interaction typically take 10 −10 s or longer to occur. This is slow in particle
physics.
*Q46.9 Answer a, b, c, and d. Protons feel all these forces; but within a nucleus the strong interaction
predominates, followed by the electromagnetic interaction.
591
592
Chapter 46
Q46.10 You can think of a conservation law as a superficial regularity which we happen to notice,
as a person who does not know the rules of chess might observe that one player’s two bishops
are always on squares of opposite colors. Alternatively, you can think of a conservation law as
identifying some stuff of which the universe is made. In classical physics one can think of both
matter and energy as fundamental constituents of the world. We buy and sell both of them. In
classical physics you can also think of linear momentum, angular momentum, and electric charge
as basic stuffs of which the universe is made. In relativity we learn that matter and energy are
not conserved separately, but are both aspects of the conserved quantity relativistic total energy.
Discovered more recently, four conservation laws appear equally general and thus equally
fundamental: Conservation of baryon number, conservation of electron-lepton number,
conservation of tau-lepton number, and conservation of muon-lepton number. Processes involving
the strong force and the electromagnetic force follow conservation of strangeness, charm,
bottomness, and topness, while the weak interaction can alter the total S, C, B, and T quantum
numbers of an isolated system.
Q46.11 No. Antibaryons have baryon number –1, mesons have baryon number 0, and baryons have
baryon number +1. The reaction cannot occur because it would not conserve baryon number,
unless so much energy is available that a baryon-antibaryon pair is produced.
Q46.12 The Standard Model consists of quantum chromodynamics (to describe the strong
interaction) and the electroweak theory (to describe the electromagnetic and weak interactions).
The Standard Model is our most comprehensive description of nature. It fails to unify the two
theories it includes, and fails to include the gravitational force. It pictures matter as made of six
quarks and six leptons, interacting by exchanging gluons, photons, and W and Z bosons.
Q46.13 (a) and (b) All baryons and antibaryons consist of three quarks. (c) and (d) All mesons and
1
antimesons consist of two quarks. Since quarks have spin quantum number
and can be spin-up
2
or spin-down, it follows that the three-quark baryons must have a half-integer spin, while the
two-quark mesons must have spin 0 or 1.
Q46.14 Each flavor of quark can have colors, designated as red, green, and blue. Antiquarks are colored
antired, antigreen, and antiblue. A baryon consists of three quarks, each having a different color.
By analogy to additive color mixing we call it colorless. A meson consists of a quark of one color
and antiquark with the corresponding anticolor, making it colorless as a whole.
Q46.15 The electroweak theory of Glashow, Salam, and Weinberg predicted the W + , W− , and
Z particles. Their discovery in 1983 confirmed the electroweak theory.
*Q46.16 Answer (e). Both trials conserve momentum. In the first trial all of the kinetic energy K1 + K2 = 2K1
is converted into internal energy. In the second trial we end up with a glob of twice the mass
moving at half the speed, so it has half the kinetic energy of one original clay ball, K1/2. Energy
K1/2 is converted into internal energy, one-quarter of that converted in trial one.
Q46.17 Hubble determined experimentally that all galaxies outside the Local Group are moving away
from us, with speed directly proportional to the distance of the galaxy from us.
*Q46.18 Answer c, b, d, e, a, f, g. The temperature corresponding to b is on the order of 1013 K. That for
hydrogen fusion d is on the order of 107 K. A fully ionized plasma can be at 104 K. Neutral atoms
can exist at on the order of 3 000 K, molecules at 1 000 K, and solids at on the order of 500 K.
Q46.19 Before that time, the Universe was too hot for the electrons to remain in any sort of stable orbit
around protons. The thermal motion of both protons and electrons was too rapid for them to be in
close enough proximity for the Coulomb force to dominate.
Particle Physics and Cosmology
593
*Q46.20 Answer (a). The vast gulfs not just between stars but between galaxies and especially between
clusters, empty of ordinary matter, are important to bring down the average density of the
Universe. We can estimate the average density defined for the Solar System as the mass of the
sun divided by the volume of a lightyear-size sphere around it:
2 × 10 30 kg
= 6 × 10 −20 kg/m 3 = 6 × 10 −23 g/cm 3 , ten million times larger than the critical
16
3
4
π
(
2
×
10
m)
3
density 3H 2/8p G = 6 × 10–30 g/cm3.
Q46.21 The Universe is vast and could on its own terms get along very well without us. But as the cosmos is immense, life appears to be immensely scarce, and therefore precious. We must do our
work, growing corn to feed the hungry while preserving our planet for future generations and
preserving future possibilities for the universe. One person has singular abilities and opportunities
for effort, faithfulness, generosity, honor, curiosity, understanding, and wonder. His or her place
is to use those abilities and opportunities, unique in all the Universe.
SOLUTIONS TO PROBLEMS
Section 46.1
The Fundamental Forces in Nature
Section 46.2
Positrons and Other Antiparticles
P46.1
Assuming that the proton and antiproton are left nearly at rest after they are produced, the energy
E of the photon must be
E = 2 E0 = 2 ( 938.3 MeV) = 1 876.6 MeV = 3.00 × 10 −10 J
Thus,
P46.2
E = hf = 3.00 × 10 −10 J
f=
3.00 × 10 −10 J
= 4.53 × 10 23 Hz
6.626 × 10 −34 J ⋅ s
λ=
c 3.00 × 10 8 m s
=
= 6.62 × 10 −16 m
f
4.53 × 10 23 Hz
The half-life of
14
O is 70.6 s, so the decay constant is λ =
ln 2
ln 2
=
= 0.009 82 s −1 .
T1 2 70.6 s
The number of 14O nuclei remaining after five minutes is
− 0.009 82 s−1 ) ( 300 s)
N = N 0 e− λ t = (1010 ) e (
= 5.26 × 10 8
The number of these in one cubic centimeter of blood is
⎞
⎛
⎛ 1.00 cm 3 ⎞
1.00 cm 3
8
N′ = N ⎜
= 2 .63 × 10 5
⎟ = ( 5.26 × 10 ) ⎜
3⎟
⎝ total volume of blood ⎠
⎝ 2000 cm ⎠
and their activity is
*P46.3
(a)
R = λ N ′ = ( 0.009 82 s−1 ) ( 2 .63 × 10 5 ) = 2 .58 × 10 3 Bq
~10 3 Bq
The rest energy of a total of 6.20 g of material is converted into energy of electromagnetic
radiation:
E = mc 2 = 6.20 × 10 −3 kg ( 3 × 10 8 m s) = 5.58 × 1014 J
2
(b)
⎛ $0.14 ⎞ ⎛ k ⎞ ⎛ W ⎞ ⎛ 1 h ⎞
7
5.58 × 1014 J = 5.58 × 1014 J ⎜
⎟ = $2.17 × 10
⎟⎜ ⎟⎜
⎟⎜
⎝ kWh ⎠ ⎝ 1 000 ⎠ ⎝ J s ⎠ ⎝ 3 600 s ⎠
All from two cents’ worth of stuff.
594
P46.4
Chapter 46
The minimum energy is released, and hence the minimum frequency photons are produced, when
the proton and antiproton are at rest when they annihilate.
That is, E = E0 and K = 0. To conserve momentum, each photon must carry away one-half the
energy.
2E
Thus
Emin = 0 = E0 = 938.3 MeV = hfmin
2
Thus,
fmin =
λ=
P46.5
(938.3 MeV) (1.60 × 10−13
(6.626 × 10
c
fmin
=
−34
J ⋅ s)
J MeV)
= 2.27 × 10 23 Hz
3.00 × 10 8 m s
= 1.32 × 10 −15 m
2.27 × 10 23 Hz
In γ → p+ + p − ,
we start with energy
2.09 GeV
we end with energy
938.3 MeV + 938.3 MeV + 95.0 MeV + K 2
where K 2 is the kinetic energy of the second proton.
K 2 = 118 MeV
Conservation of energy for the creation process gives
Section 46.3
P46.6
Mesons and the Beginning of Particle Physics
The reaction is
muon-lepton number before reaction:
electron-lepton number before reaction:
µ + + e− → ν + ν
(−1) + (0) = −1
(0) + (1) = 1
Therefore, after the reaction, the muon-lepton number must be –1. Thus, one of the neutrinos
must be the anti-neutrino associated with muons, and one of the neutrinos must be the neutrino
associated with electrons:
νµ
νe
µ + + e− → ν µ + ν e
Then
P46.7
and
The creation of a virtual Z 0 boson is an energy fluctuation ∆E = 91 × 10 9 eV. It can last no
longer than ∆t =
and move no farther than
2 ∆E
6.626 × 10 −34 J ⋅ s) ( 3.00 × 10 8 m s) ⎛
⎞
(
1 eV
hc
c ( ∆t ) =
=
⎜
⎟
−19
9
⎝ 1.60 × 10 J ⎠
4 π ∆E
4 π ( 91 × 10 eV)
= 1.06 × 10 −18 m = ~10 −18 m
.
Particle Physics and Cosmology
*P46.8
(a)
595
The particle’s rest energy is mc2. The time interval during which a virtual particle of this
mass could exist is at most ∆t in ∆E ∆t = = mc2∆t. The distance it could move is at most
2
−34
(6.626 × 10 J ⋅ s) (2.998 × 108 m/s) = 1.240 × 10−6 eV ⋅ m = 1240 eV ⋅ nm
c
c∆t =
=
2 mc 2
4 π mc 2
4 π mc 2
4 π mc 2 (1.602 × 10 −19 J/eV)
=
98.7 eV ⋅ nm
mc 2
According to Yukawa’s line of reasoning, this distance is the range of a force that could be
associated with the exchange of virtual particles of this mass.
P46.9
(b)
The range is inversely proportional to the mass of the field particle.
(c)
Our rule describes the electromagnetic, weak, and gravitational interactions.
For the electromagnetic and gravitational interactions, we take the limiting form of the
rule with infinite range and zero mass for the field particle. For the weak interaction,
98.7 eV ⋅ nm/90 GeV ≈ 10–18 m = 10–3 fm, in agreement with the tabulated information.
For the strong interaction, we do not have a separately measured mass for a gluon, so we
cannot say that this rule defines the range.
(d)
98.7 eV⋅nm/938.3 MeV ≈ 10–1–9–6 m ~10–16 m
From Table 46.2 in the chapter text M π 0 = 135 MeV c 2 .
Therefore,
and
P46.10
Eγ = 67.5 MeV for each photon
p=
Eγ
f=
Eγ
c
h
= 67.5 MeV c
= 1.63 × 10 22 Hz
The time interval for a particle traveling with the speed of light to travel a distance of 3 × 10 −15 m is
∆t =
d 3 × 10 −15 m
=
= ~10 −23 s
v 3 × 10 8 m s
596
P46.11
Chapter 46
(a)
∆E = ( mn − m p − me ) c 2
From the table of isotopic masses in Chapter 44,
∆E = (1.008 665 − 1.007 825 )( 931.5 ) = 0.782 MeV
(b)
Assuming the neutron at rest, momentum conservation for the decay process implies
p p = pe . Relativistic energy for the system is conserved
(m c )
2 2
p
(938.3)
Since p p = pe ,
Solving the algebra gives
2
+ p 2p c 2 +
+ ( pc )2 +
(m c )
2 2
e
+ pe2 c 2 = mn c 2
(0.511)
2
+ ( pc )2 = 939.6 MeV
pc = 1.19 MeV
If pe c = γ me ve c = 1.19 MeV, then
Solving,
γ ve 1.19 MeV
x
=
=
= 2.33 where
c
0.511 MeV
1 − x2
v
x 2 = (1 − x 2 ) 5.43 and x = e = 0.919
c
x=
ve
c
ve = 0.919 c
Then m p v p = γ e me ve :
vp =
−13
γ e me ve c (1.19 MeV) (1.60 × 10 J MeV)
=
mpc
(1.67 × 10−27 ) ( 3.00 × 108 m s)
v p = 3.80 × 10 5 m s = 380 km s
(c)
The electron is relativistic; the proton is not. Our criterion for answers accurate to three
significant digits is that the electron is moving at more than one-tenth the speed of light and
the proton at less than one-tenth the speed of light.
Section 46.4
P46.12
Classification of Particles
In ? + p+ → n + µ +, charge conservation requires the unknown particle to be neutral. Baryon
number conservation requires baryon number = 0. The muon-lepton number of ? must be –1.
So the unknown particle must be ν µ .
Particle Physics and Cosmology
Section 46.5
P46.13
P46.14
Le
0 + 0 → 0 +1
and
Lµ
0 + 0 → −1 + 0
(b)
π− + p → p + π+
charge
−1 + 1 → +1 + 1
(c)
p + p → p + π+
baryon number :
1+1 → 1+ 0
(d)
p+p →p+p+n
baryon number :
1+1 → 1+1+1
(e)
γ + p → n + π0
charge
0 +1 → 0 + 0
(a)
Baryon number and charge are conserved, with respective values of
and 1 + 1 = 1 + 1 in both reactions.
(b)
P46.15
Conservation Laws
p + p → µ + + e−
(a)
597
Strangeness is not conserved
0 +1 = 0 +1
in the second reaction.
(a)
π − → µ− + νµ
Lµ : 0 → 1 − 1
(b)
K+ → µ+ + νµ
Lµ : 0 → −1 + 1
(c)
ν e + p + → n + e+
Le : −1 + 0 → 0 − 1
(d)
ν e + n → p+ + e−
Le : 1 + 0 → 0 + 1
(e)
ν µ + n → p+ + µ −
Lµ : 1 + 0 → 0 + 1
(f )
µ − → e− + ν e + ν µ
Lµ : 1 → 0 + 0 + 1
and
Le : 0 → 1 − 1 + 0
P46.16
Baryon number conservation allows the first and forbids the second .
P46.17
(a)
p+ → π + + π 0
(b)
p+ + p+ → p+ + p+ + π 0
(c)
p+ + p+ → p+ + π +
(d)
π + → µ+ + νµ
This reaction can occur .
(e)
n 0 → p+ + e− + ν e
This reaction can occur .
(f )
π + → µ+ + n
Violates baryon number :
0 → 0 +1
Violates muon-lepton number :
0 → −1 + 0
Baryon number conservation is violated: 1 → 0 + 0
This reaction can occur .
Baryon number is violated:
1+1 → 1+ 0
598
P46.18
Chapter 46
Momentum conservation for the decay requires the pions to have equal speeds.
497.7 MeV
2
2
2
E = p 2 c 2 + ( mc 2 ) gives
The total energy of each is
so
(248.8 MeV)
2
= ( pc ) + (139.6 MeV)
2
pc = 206 MeV = γ mvc =
Solving,
mc 2
1 − (v c)
2
2
⎛v⎞
⎜ ⎟
⎝c⎠
⎛v⎞
pc
206 MeV
1
=
=
⎜ ⎟ = 1.48
2 ⎝c⎠
mc 2 139.6 MeV
1 − (v c)
⎛v⎞
v
= 1.48 1 − ⎜ ⎟
⎝c⎠
c
2
⎡ ⎛v⎞
⎛v⎞
⎜ ⎟ = 2.18 ⎢1 − ⎜ ⎟
⎝c⎠
⎢⎣ ⎝ c ⎠
2
and
2
2
⎤
⎛v⎞
⎥ = 2.18 − 2.18 ⎜ ⎟
⎝c⎠
⎥⎦
⎛v⎞
3.18 ⎜ ⎟ = 2.18
⎝c⎠
2
v
2.18
=
= 0.828
c
3.18
so
v = 0.828 c
and
P46.19
(a)
In the suggested reaction
p → e+ + γ
+1 → 0 + 0
We would have for baryon numbers
∆B ≠ 0, so baryon number conservation would be violated.
(b)
From conservation of momentum for the decay:
pe = pγ
Then, for the positron,
Ee2 = ( pe c ) + E02,e
becomes
Ee2 = ( pγ c ) + E02, e = Eγ2 + E02, e
From conservation of energy for the system:
E0, p = Ee + Eγ
or
Ee = E0, p − Eγ
so
Ee2 = E02, p − 2 E0, p Eγ + Eγ2
Equating this to the result from above gives
Eγ2 + E02,e = E02, p − 2 E0, p Eγ + Eγ2
or
Thus
2
2
Eγ =
E02, p − E02, e
2 E0 , p
(938.3 MeV) − (0.511 MeV)
2 ( 938.3 MeV)
2
=
Ee = E0, p − Eγ = 938.3 MeV − 469 MeV = 469 MeV
Eγ
Also,
pγ =
=
469 MeV
c
and
pe = pγ =
469 MeV
c
continued on next page
c
2
= 469 MeV
Particle Physics and Cosmology
(c)
The total energy of the positron is
Ee = 469 MeV
But,
Ee = γ E0 , e =
599
E0 , e
1 − (v c)
2
so
⎛v⎞
E
0.511 MeV
1 − ⎜ ⎟ = 0 ,e =
= 1.09 × 10 −3
⎝c⎠
469 MeV
Ee
which yields
v = 0.999 999 4 c
2
P46.20
The relevant conservation laws are
∆Le = 0
∆Lµ = 0
∆Lτ = 0
and
(a) π + → π 0 + e+ + ?
Le :
0 → 0 − 1 + Le
implies Le = 1
and we have a ν e
(b) ? + p → µ − + p + π +
Lµ :
Lµ + 0 → + 1 + 0 + 0
implies Lµ = 1
and we have a ν µ
(c) Λ 0 → p + µ − + ?
Lµ :
0 → 0 + 1 + Lµ
implies Lµ = −1
and we have a ν µ
(d) τ + → µ + + ? + ?
Lµ :
0 → − 1 + Lµ
implies Lµ = 1
and we have a ν µ
Lτ :
−1 → 0 + Lτ
implies L τ = −1 and we have a ν τ
Conclusion for (d):
Section 46.6
P46.21
Lµ = 1 for one particle, and Lτ = −1 for the other particle.
We have
νµ
and
ντ
Strange Particles and Strangeness
(a) Λ 0 → p + π −
Strangeness: −1 → 0 + 0
(strangeness is not conserved )
(b) π − + p → Λ 0 + K 0 Strangeness: 0 + 0 → −1 + 1 ( 0 = 0 and strangeness is conserved )
P46.22
(c) p + p → Λ 0 + Λ 0
Strangeness: 0 + 0 → +1 − 1 ( 0 = 0 and strangeness is conserved )
(d) π − + p → π − + Σ+
Strangeness: 0 + 0 → 0 − 1
( 0 ≠ −1 : strangeness is not conserved )
(e) Ξ− → Λ 0 + π −
Strangeness: −2 → −1 + 0
( −2 ≠ −1 so strangeness is not conserved )
(f) Ξ0 → p + π −
Strangeness: −2 → 0 + 0
( −2 ≠ 0 so strangeness is not conserved )
The ρ 0 → π + + π − decay must occur via the strong interaction.
The K S0 → π + + π − decay must occur via the weak interaction.
600
P46.23
P46.24
Chapter 46
µ − → e− + γ
Le :
0 → 1 + 0,
and
Lµ :
1→ 0
(b)
n → p + e− + ν e
Le :
0 → 0 +1+1
(c)
Λ0 → p + π 0
Strangeness:
−1 → 0 + 0,
and
charge:
0 → +1 + 0
(d)
p → e+ + π 0
Baryon number:
+1 → 0 + 0
(e)
Ξ0 → n + π 0
Strangeness:
−2 → 0 + 0
(a)
π − + p → 2η violates conservation of baryon number as 0 + 1 → 0,
(b)
K− + n → Λ0 + π −
(a)
Baryon number,
Charge,
Strangeness,
not allowed .
0 +1 → 1+ 0
−1 + 0 → 0 − 1
−1 + 0 → −1 + 0
Lepton number,
0→0
The interaction may occur via the strong interaction since all are conserved.
(c)
K− → π − + π 0
Strangeness,
−1 → 0 + 0
Baryon number,
0→0
Lepton number,
0→0
Charge,
−1 → −1 + 0
Strangeness conservation is violated by one unit, but everything else is conserved. Thus,
the reaction can occur via the weak interaction , but not the strong or electromagnetic
interaction.
(d)
Ω− → Ξ− + π 0
Baryon number,
Lepton number,
Charge,
Strangeness,
1 → 1+ 0
0→0
−1 → −1 + 0
−3 → −2 + 0
May occur by weak interaction , but not by strong or electromagnetic.
(e)
η → 2γ
Baryon number,
0→0
Lepton number,
0→0
Charge,
0→0
Strangeness,
0→0
No conservation laws are violated, but photons are the mediators of the electromagnetic
interaction. Also, the lifetime of the h is consistent with the electromagnetic interaction .
Particle Physics and Cosmology
P46.25
(a)
Ξ− → Λ 0 + µ − + ν µ
Baryon number:
+1 → +1 + 0 + 0
Charge:
−1 → 0 − 1 + 0
Le :
0 →0+0+0
Lµ :
0 → 0 +1+1
Lτ :
0 →0+0+0
Strangeness:
−2 → −1 + 0 + 0
Conserved quantities are:
(b)
B, charge, Le , and Lτ
K S0 → 2 π 0
Baryon number:
Le :
0→0
0→0
Charge:
Lµ :
0→0
0→0
Lτ :
0→0
Strangeness:
+1 → 0
Conserved quantities are:
(c)
B, charge, Le , Lµ , and Lτ
K − + p → Σ0 + n
Baryon number:
0 +1 → 1+1
Charge:
−1 + 1 → 0 + 0
Le :
0+0 →0+0
Lµ :
0+0 →0+0
Lτ :
0+0 →0+0
Strangeness:
−1 + 0 → −1 + 0
S, charge, Le , Lµ , and Lτ
Conserved quantities are:
(d)
Σ0 + Λ 0 + γ
Baryon number:
Le :
+1 → 1 + 0
0 →0+0
Charge:
Lµ :
0→0
0 →0+0
Lτ :
0 →0+0
Strangeness:
−1 → −1 + 0
Conserved quantities are:
(e)
B, S, charge, Le , Lµ , and Lτ
e+ + e − → µ + + µ −
Baryon number:
0+0 →0+0
Charge:
+1 − 1 → +1 − 1
Le :
−1 + 1 → 0 + 0
0+0 →0+0
Lµ :
0 + 0 → +1 − 1
0+0 →0+0
Lτ :
Strangeness:
Conserved quantities are:
(f)
B, S, charge, Le , Lµ , and Lτ
p + n → Λ 0 + Σ−
Baryon number:
−1 + 1 → −1 + 1
Charge:
−1 + 0 → 0 − 1
Le :
0+0 →0+0
0+0 →0+0
Lµ :
0+0 →0+0
0 + 0 → +1 − 1
Lτ :
Conserved quantities are:
Strangeness:
B, S, charge, Le , Lµ , and Lτ
601
602
P46.26
Chapter 46
(a)
K+ + p → ? + p
The strong interaction conserves everything.
Baryon number,
0 +1 → B +1
Charge,
+1 + 1 → Q + 1
Lepton numbers,
0+0 → L+0
Strangeness,
+1 + 0 → S + 0
so
so
so
so
B=0
Q = +1
Le = Lµ = Lτ = 0
S =1
The conclusion is that the particle must be positively charged, a non-baryon, with
strangeness of +1. Of particles in Table 46.2, it can only be the K + . Thus, this is an
elastic scattering process.
The weak interaction conserves all but strangeness, and ∆S = ±1.
(b)
Ω− → ? + π −
Baryon number,
+1 → B + 0
so
B =1
−1 → Q − 1
Q=0
Charge,
so
0 → L+0
Le = Lµ = Lτ = 0
Lepton numbers,
so
−3 → S + 0
∆S = 1: S = −2
Strangeness,
so
0
The particle must be a neutral baryon with strangeness of –2. Thus, it is the Ξ .
(c)
K+ → ? + µ+ + νµ
Baryon number,
Charge,
Lepton numbers,
Strangeness,
0 → B+0+0
+1 → Q + 1 + 0
Le , 0 → Le + 0 + 0
so
so
so
B=0
Q=0
Le = 0
Lµ , 0 → Lµ − 1 + 1
so
Lµ = 0
Lτ , 0 → Lτ + 0 + 0
so
Lτ = 0
1→ S+0+0
so
∆S = ±1
(for weak interaction): S = 0
The particle must be a neutral meson with strangeness = 0 ⇒ π 0 .
P46.27
Time-dilated lifetime:
T = γ T0 =
0.900 × 10 −10 s
1− v c
2
2
=
0.900 × 10 −10 s
1 − (0.960 )
2
= 3.214 × 10 −10 s
distance = 0.960 ( 3.00 × 10 8 m s) ( 3.214 × 10 −10 s) = 9.26 cm
Particle Physics and Cosmology
P46.28
(a)
pΣ+ = eBrΣ+ =
pπ + = eBrπ + =
(b)
(1.602 177 × 10
−19
5.344 288 × 10
(1.602 177 × 10
−22
−19
5.344 288 × 10
−22
C) (1.15 T)(1.99 m )
(kg ⋅ m s) (MeV c)
=
C) (1.15 T) ( 0.580 m )
(kg ⋅ m s) (MeV c)
=
603
686 MeV
c
200 MeV
c
Let φ be the angle made by the neutron’s path with the path of the Σ+ at the moment of
decay. By conservation of momentum:
pn cos φ + (199.961 581 MeV c ) cos 64.5° = 686.075 081 MeV c
∴ pn cos φ = 599.989 401 MeV c
(1)
pn sin φ = (199.961 581 MeV c ) sin 64.5° = 180.482 380 MeV c
(2)
(599.989 401 MeV c) + (180.482 380
2
pn =
From (1) and (2):
MeV c )
2
= 627 MeV c
(c)
Eπ + =
En =
( p c) + ( m
( pn c )
2
c2 ) =
2
2
π+
π+
+ ( mn c 2 ) =
2
(199.961 581 MeV) + (139.6 MeV)
2
(626.547 022 MeV) + (939.6 MeeV)
2
2
2
= 244 MeV
= 1130 MeV
EΣ+ = Eπ + + En = 243.870 445 MeV + 1129.340 219 MeV = 1 370 MeV
(d)
mΣ+ c 2 = EΣ2+ − ( pΣ+ c ) =
2
(1 373.210 664 MeV) − (686.075 081 MeV)
2
∴ mΣ+ = 1190 MeV c 2
−1 2
⎛ v2 ⎞
EΣ+ = γ mΣ+ c 2 , where γ = ⎜1 − 2 ⎟
⎝ c ⎠
Solving for v, we find v = 0.500 c .
=
1 373.210 664 MeV
= 1.154 4
1189.541 3033 MeV
2
= 1190 MeV
604
P46.29
Chapter 46
(a)
Let Emin be the minimum total energy of the bombarding particle that is needed to induce
the reaction. At this energy the product particles all move with the same velocity. The product particles are then equivalent to a single particle having mass equal to the total mass of
the product particles, moving with the same velocity as each product particle. By conservation of energy:
( m c ) + ( p c)
2 2
Emin + m2 c 2 =
3
2
(1)
3
By conservation of momentum: p3 = p1
2
∴ ( p3 c ) = ( p1c ) = Emin
− ( m1c 2 )
2
2
Emin + m2 c 2 =
Substitute (2) in (1):
2
(2)
(m c )
2 2
3
2
+ Emin
− ( m1c 2 )
2
Square both sides:
2
2
Emin
+ 2 Emin m2 c 2 + ( m2 c 2 ) = ( m3 c 2 ) + Emin
− ( m1c 2 )
2
∴ Emin =
(m
2
3
2
2
− m12 − m22 ) c 2
2 m2
∴ K min = Emin − m1c
2
(m
=
2
3
− m12 − m22 − 2 m1 m2 ) c 2
2 m2
⎡m 2 − ( m + m )2 ⎤ c 2
1
2
⎣ 3
⎦
=
2m2
Refer to Table 46.2 for the particle masses.
⎡⎣4 ( 938.3)⎤⎦ MeV2 c 2 − ⎡⎣2 ( 938.3)⎤⎦ MeV2 c 2
=
= 5.63 GeV
2 ( 938.3 MeV c 2 )
2
(b)
K min
(c)
K min =
( 497.7 + 1115.6)
2
2
(e)
2
2 ( 938.3) MeV c 2
= 768 MeV
K min
⎡⎣2 ( 938.3) + 135⎤⎦ MeV2 c 2 − ⎡⎣2 ( 938.3)⎤⎦ MeV2 c 2
=
= 280 MeV
2 ( 938.3) MeV c 2
K min
⎡ 91.2 × 10 3 2 − ⎡ 938.3 + 938.3 2 ⎤ MeV2 c 2 ⎤
)⎦
) ⎣(
⎥⎦
⎣⎢(
=
= 4.43 TeV
2 ( 938.3) MeV c 2
2
(d)
MeV2 c 2 − (139.6 + 938.3) MeV2 c 2
2
Particle Physics and Cosmology
Section 46.7
Finding Patterns in the Particles
Section 46.8
Quarks
Section 46.9
Multicolored Quarks
Section 46.10
The Standard Model
P46.30
(a)
605
The number of protons
⎛ 6.02 × 10 23 molecules ⎞ ⎛ 10 protons ⎞
26
N p = 1 000 g ⎜
⎟ = 3.34 × 10 protons
⎟⎜
⎝
⎠
molecule
18.0
g
⎝
⎠
and there are
⎛ 6.02 × 10 23 molecules ⎞ ⎛ 8 neutrons ⎞
26
Nn = 1 000 g ⎜
⎟ = 2.68 × 10 neutrons
⎟⎜
18.0 g
⎝
⎠ ⎝ molecule ⎠
So there are for electric neutrality
(b)
3.34 × 10 26 electrons
The up quarks have number
2 ( 3.34 × 10 26 ) + 2.68 × 10 26 = 9.36 × 10 26 up quarks
and there are
2 ( 2.68 × 10 26 ) + 3.34 × 10 26 = 8.70 × 10 26 down quarkss
Model yourself as 65 kg of water. Then you contain:
65 ( 3.34 × 10 26 ) ~10 28 electrons
65 ( 9.36 × 10 26 ) ~10 29 up quarks
65 ( 8.70 × 10 26 ) ~10 29 down quarks
Only these fundamental particles form your body. You have no strangeness, charm,
topness, or bottomness.
P46.31
(a)
proton
strangeness
baryon number
charge
(b)
strangeness
baryon number
charge
0
1
e
u
0
1/3
2e/3
u
0
1/3
2e/3
d
0
1/3
–e/3
total
0
1
e
neutron
u
d
d
total
0
1
0
0
1/3
2e/3
0
1/3
–e/3
0
1/3
–e/3
0
1
0
606
P46.32
Chapter 46
Quark composition of proton = uud and of neutron = udd.
Thus, if we neglect binding energies, we may write
and
mp = 2 mu + md
(1)
mn = mu + 2 md
(2)
Solving simultaneously,
mu =
we find
1
1
2 mp − mn) = ⎡⎣2 ( 938 MeV c 2 ) − 939.6 meV c 2 ⎦⎤ = 312 MeV c 2
(
3
3
and from either (1) or (2), md = 314 MeV c 2 .
P46.33
(a)
strangeness
baryon number
charge
(b)
strangeness
baryon number
charge
P46.34
K0
d
s
total
1
0
0
0
1/3
–e/3
1
–1/3
e/3
1
0
0
Λ0
u
d
s
total
–1
1
0
0
1/3
2e/3
0
1/3
–e/3
–1
1/3
–e/3
–1
1
0
In the first reaction, π − + p → K 0 + Λ 0 , the quarks in the particles are ud + uud → ds + uds. There
is a net of 1 up quark both before and after the reaction, a net of 2 down quarks both before and
after, and a net of zero strange quarks both before and after. Thus, the reaction conserves the net
number of each type of quark.
In the second reaction, π − + p → K 0 + n, the quarks in the particles are ud + uud → ds + udd.
In this case, there is a net of 1 up and 2 down quarks before the reaction but a net of 1 up,
3 down, and 1 anti-strange quark after the reaction. Thus, the reaction does not conserve the net
number of each type of quark.
P46.35
(a)
π − + p → K0 + Λ0
In terms of constituent quarks:
up quarks:
down quarks:
strange quarks:
(b)
(c)
π + + p → K + + Σ+
ud + uud → ds + uds
−1 + 2 → 0 + 1,
1 + 1 → 1 + 1,
0 + 0 → −1 + 1,
or
1→1
2→2
0→0
or
or
du + uud → us + uus
up quarks:
1 + 2 → 1 + 2,
or
3→ 3
down quarks:
−1 + 1 → 0 + 0,
or
0→0
strange quarks:
0 + 0 → −1 + 1,
or
0→0
or
or
or
1→1
1→1
1→1
K − + p → K + + K 0 + Ω−
up quarks:
down quarks:
strange quarks:
continued on next page
us + uud → us + ds + sss
−1 + 2 → 1 + 0 + 0,
0 + 1 → 0 + 1 + 0,
1 + 0 → −1 − 1 + 3,
Particle Physics and Cosmology
(d)
p + p → K0 + p + π + + ?
607
uud + uud → ds + uud + ud + ?
The quark combination of ? must be such as to balance the last equation for up, down, and
strange quarks.
up quarks:
2 + 2 = 0 + 2 +1+ ?
(has 1 u quark)
down quarks:
1+1 = 1+1 −1+ ?
(has 1 d quark)
0 + 0 = −1 + 0 + 0 + ?
(has 1 s quark)
strange quarks:
quark composition = uds = Λ or Σ
0
P46.36
0
Σ0 + p → Σ+ + γ + X
dds + uud → uds + 0 + ?
The left side has a net 3d, 2u, and 1s. The right-hand side has 1d, 1u, and 1s leaving 2d and 1u
missing.
The unknown particle is a neutron, udd.
Baryon and strangeness numbers are conserved.
P46.37
P46.38
Compare the given quark states to the entries in Tables 46.4 and 46.5:
(a)
suu = Σ+
(b)
ud = π −
(c)
sd = K 0
(d)
ssd = Ξ−
(a)
uud:
⎛ 2 ⎞ ⎛ 2 ⎞ ⎛1 ⎞
charge = ⎜− e ⎟ + ⎜− e ⎟ + ⎜ e ⎟ = −e . This is the antiproton .
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝3 ⎠
(b)
udd:
⎛ 2 ⎞ ⎛1 ⎞ ⎛1 ⎞
charge = ⎜− e ⎟ + ⎜ e ⎟ + ⎜ e ⎟ = 0 . This is the antineutron .
⎝ 3 ⎠ ⎝3 ⎠ ⎝3 ⎠
Section 46.11
The Cosmic Connection
*P46.39 We let r in Hubble’s law represent any distance.
(a)
⎞⎛
⎛ 1 ly ⎞ ⎛
⎞
m
1 yr
c
1.85 m ⎜
⎟⎜
⎟
⎟⎜
8
7
s ⋅ ly
⎝ c ⋅ 1 yr ⎠ ⎝ 3 × 10 m s ⎠ ⎝ 3.156 × 10 s ⎠
m
1
1.85 m = 1.80 × 10 −18 1.85 m = 3.32 × 10 −18 m s
= 17 × 10 −3
s ⋅ 9.47 × 1015 m
s
v = Hr = 17 × 10 −3
This is unobservably small.
(b)
1
v = Hr = 1.80 × 10 −18 3.84 × 10 8 m = 6.90 × 10 −10 m s
s
again too small to measure
608
P46.40
Chapter 46
Section 39.4 says
fobserver = fsource
1 + va c
1 − va c
The velocity of approach, va , is the negative of the velocity of mutual recession: va = −v.
c c 1− v c
=
λ′ λ 1+ v c
Then,
P46.41
(a)
λ′ = λ
1+ v c
1− v c
λ′ = λ
and
510 nm = 434 nm
1+ v c
= 1.381
1− v c
v
v
1 + = 1.381 − 1.381
c
c
v
= 0.160
or
c
2.38
(b)
v = HR :
R=
(a)
λn′ = λn
1+ v c
1− v c
1+ v c
1− v c
1.18 2 =
P46.42
1+
v
= 0.381
c
v = 0.160 c = 4.80 × 10 7 m s
v
4.80 × 10 7 m s
=
= 2.82 × 10 9 ly
H 17 × 10 −3 m s ⋅ ly
1+ v c
2
= ( Z + 1)
1− v c
1+ v c
= ( Z + 1) λn
1− v c
⎛v⎞ 2
2
⎜ ⎟ ( Z + 2 Z + 2) = Z + 2 Z
⎝c⎠
⎛v⎞
v
2
2
= ( Z + 1) − ⎜ ⎟ ( Z + 1)
⎝c⎠
c
⎛ Z 2 + 2Z ⎞
v = c⎜ 2
⎝ Z + 2 Z + 2 ⎟⎠
(b)
P46.43
R=
v = HR
v
c ⎛ Z 2 + 2Z ⎞
=
H
H ⎜⎝ Z 2 + 2 Z + 2 ⎟⎠
(1.7 × 10
H=
m s)
−2
ly
1+ v c
= 590 (1.000 113 3) = 590.07 nm
1− v c
(a)
v ( 2.00 × 10 6 ly ) = 3.4 × 10 4 m s
λ′ = λ
(b)
v ( 2.00 × 10 8 ly ) = 3.4 × 10 6 m s
λ ′ = 590
1 + 0.011 33
= 597 nm
1 − 0.011 33
(c)
v ( 2.00 × 10 9 ly ) = 3.4 × 10 7 m s
λ ′ = 590
1 + 0.113 3
= 661 nm
1 − 0.113 3
Particle Physics and Cosmology
*P46.44 (a)
(b)
609
What we can see is limited by the finite age of the Universe and by the finite speed of light.
We can see out only to a look-back time equal to a bit less than the age of the Universe.
Every year on your birthday the Universe also gets a year older, and light now in
transit from still more distant objects arrives at Earth. So the radius of the visible Universe
expands at the speed of light, which is dr/dt = c = 1 ly/yr.
The volume of the visible section of the Universe is (4/3)p r 3 where
r = 13.7 billion light-years. The rate of volume increase is
⎛
3 × 10 8 ms 3.156 × 10 7 s ⎞
dV d 43 π r 3 4
dr
8
=
= 3 π 3r 2
= 4 π r 2 c = 4 π ⎜13.7 × 10 9 ly
⎟ 3 × 10
1 ly
dt
dt
dt
⎝
⎠
2
m
s
= 6.34 × 1061 m3/s
*P46.45 (a)
The volume of the sphere bounded by the Earth’s orbit is
3
4 3 4
π r = π (1.496 × 1011 m ) = 1.40 × 10 34 m 3
3
3
m = ρV = 6 × 10 −28 kg m 3 1.40 × 10 34 m 3 = 8.41 × 10 6 kg
V=
(b)
By Gauss’s law, the dark matter would create a gravitational field acting on the Earth to
accelerate it toward the Sun. It would shorten the duration of the year in the same way that
8.41 × 10 6 kg of extra material in the Sun would. This has the fractional effect of
8.41 × 10 6 kg
= 4.23 × 10 −24 of the mass of the Sun. It is immeasurably small .
1.99 × 10 30 kg
P46.46
(a)
Wien’s law:
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
=
= 1.06 × 10 −3 m = 1.06 mm
T
2.73 K
.
This is a microwave .
Thus,
(b)
P46.47
λmaxT = 2.898 × 10 −3 m ⋅ K
λmax =
We assume that the fireball of the Big Bang is a black body.
I = eσ T 4 = (1) ( 5.67 × 10 −8 W m 2 ⋅ K 4 ) ( 2.73 K ) = 3.15 × 10 −6 W m 2
4
As a bonus, we can find the current power of direct radiation from the Big Bang in the portion of
the Universe observable to us. If it is fourteen billion years old, the fireball is a perfect sphere of
radius fourteen billion light years, centered at the point halfway between your eyes:
⎛ 3 × 10 8 m s ⎞
2
7
⎟ ( 3.156 × 10 s yr )
⎝ 1 ly yr ⎠
P = IA = I (4 π r 2 ) = ( 3.15 × 10−6 W m 2 ) ( 4 π ) (14 × 10 9 ly ) ⎜
2
P = 7 × 10 47 W
2
610
P46.48
Chapter 46
The density of the Universe is
⎛ 3H 2 ⎞
ρ = 1.20 ρ c = 1.20 ⎜
⎟
⎝ 8π G ⎠
Consider a remote galaxy at distance r. The mass interior to the sphere below it is
⎛ 3 H 2 ⎞ ⎛ 4 3 ⎞ 0.600 H 2 r 3
4
M = ρ ⎛ π r 3 ⎞ = 1.20 ⎜
πr =
⎝3
⎠
⎠
⎝ 8π G ⎟⎠ ⎝ 3
G
both now and in the future when it has slowed to rest from its current speed v = Hr. The energy
of this galaxy-sphere system is constant as the galaxy moves to apogee distance R:
GmM
GmM
1
= 0−
mv 2 −
r
R
2
so
Gm ⎛ 0.600 H 2 r 3 ⎞
Gm ⎛ 0.600 H 2 r 3 ⎞
1
mH 2 r 2 −
0
=
−
⎟⎠
⎟⎠
r ⎜⎝
G
R ⎜⎝
G
2
r
so
R = 6.00 r
R
The Universe will expand by a factor of 6.00 from its current dimensions.
−0.100 = −0.600
P46.49
(a)
kBT ≈ 2 m p c 2
so
(b)
(a)
(b)
2 m p c2
kB
=
2 ( 938.3 MeV) ⎛ 1.60 × 10 −13 J ⎞
13
⎜
⎟ ~ 10 K
(1.38 × 10−23 J K ) ⎝ 1 MeV ⎠
kBT ≈ 2 me c 2
so
P46.50
T≈
T≈
2 ( 0.511 MeV) ⎛ 1.60 × 10 −13 J ⎞
2 me c 2
10
=
⎜
⎟ ~ 10 K
kB
(1.38 × 10−23 J K ) ⎝ 1 MeV ⎠
The Hubble constant is defined in v = HR. The distance R between any two far-separated
objects opens at constant speed according to R = v t . Then the time t since the Big Bang is
found from
1
v = H vt
1 = Ht
t=
H
⎛ 3 × 10 8 m s ⎞
1
1
10
=
⎜
⎟ = 1.76 × 10 yr = 17.6 billion years
H 17 × 10 −3 m s ⋅ ly ⎝ 1 ly yr ⎠
Particle Physics and Cosmology
P46.51
(a)
611
Consider a sphere around us of radius R large compared to the size of galaxy clusters. If the
matter M inside the sphere has the critical density, then a galaxy of mass m at the surface of
the sphere is moving just at escape speed v according to
1
GMm
mv 2 −
=0
2
R
The energy of the galaxy-sphere system is conserved, so this equation is true throughout the
dR
history of the Universe after the Big Bang, where v =
. Then
dt
2
R
T
⎛ dR ⎞ 2 GM
dR
R dR = 2 GM ∫ d t
= R−1/2 2 GM
⎜ ⎟ =
∫
0
0
dt
R
⎝dt ⎠
K + Ug = 0
R 3/ 2
32
R
= 2 GM t
2 3/ 2
R = 2 GM T
3
T
0
0
T=
2GM
=v
R
From above,
(b)
P46.52
R
2 R3 2
2
=
3 2 GM 3 2GM R
2R
3v
so
T=
Now Hubble’s law says
v = HR
So
T=
T=
3 (17 × 10
2
−3
2 R
2
=
3 HR 3H
⎛ 3 × 10 8 m s ⎞
10
⎜
⎟ = 1.18 × 10 yr = 11.8 billion years
1
ly
yr
m s ⋅ ly ) ⎝
⎠
In our frame of reference, Hubble’s law is exemplified by v1 = H R1 and v 2 = H R 2 . From
these we may form the equations −v1 = − H R1 and v 2 − v1 = H R 2 − R1 . These equations
express Hubble’s law as seen by the observer in the first galaxy cluster, as she looks at us to
find −v1 = H −R1 and as she looks at cluster two to find v 2 − v1 = H R 2 − R1 .
(
(
Section 46.12
*P46.53 (a)
(b)
(c)
)
)
(
)
Problems and Perspectives
L=
G
=
c3
(1.055 × 10
− 34
J ⋅ s) ( 6.67 × 10 −11 N ⋅ m 2 kg 2 )
( 3.00 × 10
8
m s)
3
= 1.61 × 10 −35 m
L 1.61 × 10 − 35 m
=
= 5.38 × 10 −44 s ,
c 3.00 × 10 8 m s
which is approximately equal to the duration of the ultra-hot epoch.
The Planck time is given as T =
Yes. The uncertainty principle foils any attempt at making observations of things when the
age of the Universe was less than the Planck time. The opaque fireball of the Big Bang,
measured as the cosmic microwave background radiation, prevents us from receiving
visible light from things before the Universe was a few hundred thousand years old. Walls
of more profound fire hide all information from still earlier times.
612
Chapter 46
Additional Problems
P46.54
We find the number N of neutrinos:
10 46 J = N ( 6 MeV) = N ( 6 × 1.60 × 10 −13 J )
N = 1.0 × 10 58 neutrinos
The intensity at our location is
⎛
⎞
1.0 × 10 58
1 ly
N
N
−2
14
⎜
⎟
=
=
2
2
8
7
⎜ 3.00 × 10 m s 3.16 × 10 s ⎟ = 3.1 × 10 m
5
A 4π r
)(
)⎠
4 π (1.7 × 10 ly ) ⎝ (
2
The number passing through a body presenting 5 000 cm 2 = 0.50 m 2
⎛
1 ⎞
14
0.50 m 2 ) = 1.5 × 1014
⎜ 3.1 × 10
2 ⎟(
⎝
⎠
m
is then
~ 1014
or
P46.55
A photon travels the distance from the Large Magellanic Cloud to us in 170 000 years.
The hypothetical massive neutrino travels the same distance in 170 000 years plus 10 seconds:
c (170 000 yr ) = v (170 000 yr + 10 s)
v
170 000 yr
1
1
=
=
=
c 170 000 yr + 10 s 1 + 10 s ⎡(1.7 × 10 5 yr ) ( 3.156 × 10 7 s yr ) ⎤
1 + 1.86 × 10 −12
⎣
⎦
{
}
For the neutrino we want to evaluate mc 2 in E = γ mc 2:
v2
E
1
mc = = E 1 − 2 = 10 MeV 1 −
= 10 MeV
2
γ
c
(1 + 1.86 × 10−12 )
2
mc 2 ≈ 10 MeV
2 (1.86 × 10 −12 )
1
(1 + 1.86 × 10 ) − 1
(1 + 1.86 × 10 )
−12 2
−12 2
= 10 MeV (1.93 × 10 −6 ) = 19 eV
Then the upper limit on the mass is
m=
P46.56
P46.57
19 eV
c2
or
m=
⎞
19 eV ⎛
u
= 2.1 × 10 −8 u
⎜
2
6
2 ⎟
c ⎝ 931.5 × 10 eV c ⎠
(a)
π − + p → Σ+ + π 0
is forbidden by charge conservation
(b)
µ− → π − + νe
is forbidden by energy conservation
(c)
p → π+ + π+ + π−
is forbidden by baryon number conservation
The total energy in neutrinos emitted per second by the Sun is:
2
( 0.4 ) ⎡⎣4π (1.5 × 1011 ) ⎤⎦ W = 1.1 × 10 23 W
Over 10 9 years, the Sun emits 3.6 × 10 39 J in neutrinos. This represents an annihilated mass
m c 2 = 3.6 × 10 39 J
m = 4.0 × 10 22 kg
About 1 part in 50 000 000 of the Sun’s mass, over 10 9 years, has been lost to neutrinos.
Particle Physics and Cosmology
P46.58
613
p + p → p +π+ + X
The protons each have 70.4 MeV of kinetic energy. In accord with conservation of momentum for
the collision, particle X has zero momentum and thus zero kinetic energy. Conservation of system
energy then requires
(
) (
M p c2 + M π c2 + M X c2 = M p c2 + K p + M p c2 + K p
)
M X c 2 = M p c 2 + 2 K p − M π c 2 = 938.3 MeV + 2 ( 70.4 MeV ) − 139.6 MeV = 939.5 MeV
X must be a neutral baryon of rest energy 939.5 MeV. Thus X is a neutron .
P46.59
(a)
If 2N particles are annihilated, the energy released is 2 Nmc 2 . The resulting photon
E 2 Nmc 2
momentum is p = =
= 2 Nmc. Since the momentum of the system is conserved,
c
c
the rocket will have momentum 2Nmc directed opposite the photon momentum.
p = 2 Nmc
(b)
Consider a particle that is annihilated and gives up its rest energy mc 2 to another particle
which also has initial rest energy mc 2 (but no momentum initially).
E 2 = p 2 c 2 + ( mc 2 )
2
2
2 2
2
Thus ( 2 mc ) = p c + ( mc )
2
2
where p is the momentum the second particle acquires as a result of the annihilation of the
2
2
2
first particle. Thus 4 ( mc 2 ) = p 2 c 2 + ( mc 2 ) , p 2 = 3 ( mc 2 ) . So p = 3mc.
N
N
protons and
antiprotons). Thus the total
2
2
momentum acquired by the ejected particles is 3Nmc, and this momentum is imparted to
the rocket.
This process is repeated N times (annihilate
p = 3 Nmc
(c)
P46.60
Method (a) produces greater speed since 2 Nmc > 3Nmc.
By relativistic energy conservation in the reaction,
Eγ + me c 2 =
By relativistic momentum conservation for the system,
Eγ
Dividing (2) by (1),
X=
Subtracting (2) from (1),
me c 2 =
Solving, 1 =
3 − 3X
1− X
2
and X =
c
4
so Eγ = 4 me c 2 = 2.04 MeV
5
=
3me c 2
1 − v2 c2
3me v
(2)
1 − v2 c2
Eγ
Eγ + me c
2
=
3me c 2
1− X 2
(1)
v
c
−
3me c 2 X
1− X 2
614
P46.61
Chapter 46
mΛ c 2 = 1115.6 MeV
Λ0 → p + π −
m p c 2 = 938.3 MeV
mπ c 2 = 139.6 MeV
The difference between starting rest energy and final rest energy is the kinetic energy of the
products.
K p + K π = 37.7 MeV
and
p p = pπ = p
Applying conservation of relativistic energy to the decay process, we have
⎡
⎣⎢
(938.3)
2
2
⎤ ⎡
⎤
+ p 2 c 2 − 938.3⎥ + ⎢ (139.6 ) + p 2 c 2 − 139.6⎥ = 37.7 MeV
⎦ ⎣
⎦
Solving the algebra yields
pπ c = p p c = 100.4 MeV
Then,
P46.62
Kp =
(m c )
Kπ =
(139.6 )2 + (100.4 )2 − 139.6 = 32.3 MeV
2 2
p
+ (100.4 ) − m p c 2 = 5.35 MeV
2
p + p → p +n +π+
The total momentum is zero before the reaction. Thus, all three particles present after the
reaction may be at rest and still conserve system momentum. This will be the case when the
incident protons have minimum kinetic energy. Under these conditions, conservation of energy
for the reaction gives
(
)
2 m p c 2 + K p = m p c 2 + mn c 2 + mπ c 2
so the kinetic energy of each of the incident protons is
Kp =
P46.63
mn c 2 + mπ c 2 − m p c 2
2
=
(939.6 + 139.6 − 938.3) MeV =
2
70.4 MeV
Σ0 → Λ0 + γ
From Table 46.2, mΣ = 1192.5 MeV c 2
and
mΛ = 1115.6 MeV c 2
Conservation of energy in the decay requires
(
)
E0, Σ = Eo, Λ + K Λ + Eγ
or
⎛
p 2 ⎞
1192.5 MeV = ⎜1115.6 MeV + Λ ⎟ + Eγ
2 mΛ ⎠
⎝
System momentum conservation gives pΛ = pγ , so the last result may be written as
⎛
p2 ⎞
1192.5 MeV = ⎜⎜1115.6 MeV + γ ⎟⎟ + Eγ
2 mΛ ⎠
⎝
or
⎛
p 2 c2 ⎞
1192.5 MeV = ⎜⎜1115.6 MeV + γ 2 ⎟⎟ + Eγ
2 mΛ c ⎠
⎝
Recognizing that
mΛ c 2 = 1115.6 MeV
we now have
1192.5 MeV = 1115.6 MeV +
Solving this quadratic equation gives
Eγ = 74.4 MeV
and
pγ c = Eγ
Eγ 2
2 (1115.6 MeV)
+ Eγ
Particle Physics and Cosmology
P46.64
615
The momentum of the proton is
qBr = (1.60 × 10 −19 C) ( 0.250 kg C ⋅ s)(1.33 m )
p p = 5.32 × 10 −20 kg ⋅ m s
cp p = 1.60 × 10 −11 kg ⋅ m 2 s 2 = 1.60 × 10 −11 J = 99.8 MeV
p p = 99.8 MeV c
Therefore
E p = E02 + ( cp ) =
2
The total energy of the proton is
(938.3) + (99.8)
2
2
= 944 MeV
For the pion, the momentum qBr is the same (as it must be from conservation of momentum in a
2-particle decay).
pπ = 99.8 MeV c
E0π = 139.6 MeV
Eπ = E02 + ( cp ) =
(139.6 )2 + ( 99.8)2 = 172 MeV
2
Thus Etotal after = Etotal before = Rest energy
Rest energy of unknown particle = 944 MeV + 172 MeV = 1116 MeV
(This is a Λ 0 particle!)
Mass = 1116 MeV c 2 .
P46.65
π − → µ− + νµ :
From the conservation laws for the decay,
mπ c 2 = 139.6 MeV = Eµ + Eν
P46.66
( )
[1]
and pµ = pν , Eν = pν c :
Eµ2 = pµ c + (105.7 MeV ) = ( pν c ) + (105.7 MeV )
or
Eµ2 − Eν2 = (105.7 MeV)
Since
Eµ + Eν = 139.6 MeV
and
(E
then
Eµ − Eν =
Subtracting [3] from [1],
2 Eν = 59.6 MeV
µ
2
2
2
)(
2
2
[2]
[1]
)
+ Eν Eµ − Eν = (105.7 MeV )
(105.7 MeV )2
139.6 MeV
2
= 80.0
and
[3]
Eν = 29.8 MeV
The expression e− E kBT dE gives the fraction of the photons that have energy between E and
E + dE. The fraction that have energy between E and infinity is
∞
∞
∫ e− E kBT dE
E
∞
∫e
− E kBT
dE
0
=
∫e
E
∞
∫e
− E kBT
(−dE
kBT )
=
− E kBT
(−dE
kBT )
e− E kBT
e
∞
E
− E kBT ∞
= e− E kBT
0
0
We require T when this fraction has a value of 0.0100 (i.e., 1.00%)
and
E = 1.00 eV = 1.60 × 10 −19 J
Thus,
− 1.60 ×10 −19 J ) (1.38×10 −23 J K )T
0.010 0 = e (
or ln ( 0.010 0 ) = −
[2]
1.60 × 10 −19 J
1.16 × 10 4 K
=
−
giving T = 2.52 × 10 3 K
T
(1.38 × 10 −23 J K ) T
616
P46.67
Chapter 46
(a)
This diagram represents the annihilation of an
electron and an antielectron. From charge and
lepton-number conservation at either vertex, the
exchanged particle must be an electron, e − .
(b)
This is the tough one. A neutrino collides with a
neutron, changing it into a proton with release of a
muon. This is a weak interaction. The exchanged
particle has charge +e and is a W + .
FIG. P46.67
P46.68
(a)
The mediator of this weak interaction is a Z 0 boson .
(b)
The Feynman diagram shows a down quark and its
antiparticle annihilating each other. They can produce a
particle carrying energy, momentum, and angular
momentum, but zero charge, zero baryon number, and, it
may be, no color charge. In this case the product particle
is a photon .
FIG. P46.68
For conservation of both energy and momentum in the
collision we would expect two photons; but momentum need not be strictly conserved,
according to the uncertainty principle, if the photon travels a sufficiently short distance
before producing another matter-antimatter pair of particles, as shown in Figure P46.68.
Depending on the color charges of the d and d quarks, the ephemeral particle could also be
a gluon , as suggested in the discussion of Figure 46.13(b).
P46.69
(a)
At threshold, we consider a photon and a proton colliding head-on to produce a proton and
a pion at rest, according to p + γ → p + π 0 . Energy conservation gives
mp c 2
2
1− u c
2
+ Eγ = mp c 2 + mπ c 2
mp u
Momentum conservation gives
1− u c
2
2
−
Eγ
c
= 0.
Combining the equations, we have
mp c 2
1− u c
2
2
+
mp c 2
u
= mp c 2 + mπ c 2
c
1− u c
2
938.3 MeV (1 + u c )
(1 − u c) (1 + u c)
(b)
so
u
= 0.134
c
and
Eγ = 127 MeV
λmaxT = 2.898 mm ⋅ K
λmax =
continued on next page
2.898 mm ⋅ K
= 1.06 mm
2.73 K
2
= 938.3 MeV + 135.0 MeV
Particle Physics and Cosmology
(c)
(d)
617
hc 1 240 eV ⋅ 10 −9 m
=
= 1.17 × 10 −3 eV
−3
λ
1.06 × 10 m
u′
In the primed reference frame, the proton is moving to the right at
= 0.134 and
c
8
the photon is moving to the left with hf ′ = 1.27 × 10 eV. In the unprimed frame,
hf = 1.17 × 10 −3 eV. Using the Doppler effect equation from Section 39.4, we have for the
speed of the primed frame
Eγ = hf =
1.27 × 10 8 =
1+ v c
1.17 × 10 −3
1− v c
v
= 1 − 1.71 × 10 −22
c
Then the speed of the proton is given by
u u′ c + v c
0.134 + 1 − 1.71 × 10 −22
=
=
= 1 − 1.30 × 10 −22
c 1 + u ′v c 2 1 + 0.134 (1 − 1.71 × 10 −22 )
And the energy of the proton is
m p c2
1− u c
2
2
=
938.3 MeV
1 − (1 − 1.30 × 10
)
−22 2
= 6.19 × 1010 × 938.3 × 10 6 eV = 5.81 × 1019 eV
ANSWERS TO EVEN PROBLEMS
P46.2
~103 Bq
P46.4
2.27 × 10 23 Hz; 1.32 fm
P46.6
ν µ and ν e
P46.8
(b) The range is inversely proportional to the mass of the field particle. (c) Our rule describes
the electromagnetic, weak, and gravitational interactions. For the electromagnetic and gravitational interactions, we take the limiting form of the rule with infinite range and zero mass for the
field particle. For the weak interaction, 98.7 eV ⋅ nm/90 GeV ≈ 10–18 m = 10–3 fm, in agreement
with the tabulated information. (d) ~10–16 m
P46.10
~10 −23 s
P46.12
νµ
P46.14
(b) The second violates strangeness conservation.
P46.16
The second violates conservation of baryon number.
P46.18
0.828c
P46.20
(a) ν e
P46.22
See the solution.
P46.24
(a) not allowed; violates conservation of baryon number (b) strong interaction
interaction (d) weak interaction (e) electromagnetic interaction
P46.26
(a) K+ (b) Ξ0
(b) ν µ
(c) ν µ
(c) p 0
(d) ν µ + ν τ
(c) weak
618
P46.28
Chapter 46
(a)
686 MeV
200 MeV
and
c
c
(b) 627 MeV c
(c) 244 MeV, 1130 MeV, 1 370 MeV
(d) 1190 MeV c 2, 0.500 c
P46.30
(a) 3.34 × 1026 e−, 9.36 × 1026 u, 8.70 × 1026 d (b) ~1028 e−, ~1029 u, ~1029 d. I have zero
strangeness, charm, topness, and bottomness.
P46.32
mu = 312 MeV/c2
P46.34
See the solution.
P46.36
a neutron, udd
P46.38
(a) –e, antiproton b) 0, antineutron
P46.40
See the solution.
P46.42
⎛ Z 2 + 2Z ⎞
(a) v = c ⎜ 2
⎝ Z + 2 Z + 2 ⎟⎠
md = 314 MeV/c2
(b)
c ⎛ Z 2 + 2Z ⎞
⎜
⎟
H ⎝ Z 2 + 2Z + 2 ⎠
P46.44
(a) What we can see is limited by the finite age of the Universe and by the finite speed of light.
We can see out only to a look-back time equal to a bit less than the age of the Universe. Every
year on your birthday the Universe also gets a year older, and light now in transit from still
more distant objects arrives at Earth. So the radius of the visible Universe expands at the
speed of light, which is 1 ly/yr. (b) 6.34 × 1061 m3/s
P46.46
(a) 1.06 mm
P46.48
6.00
P46.50
(a) See the solution.
P46.52
See the solution.
P46.54
~1014
P46.56
(a) charge (b) energy
P46.58
neutron
P46.60
2.04 MeV
P46.62
70.4 MeV
P46.64
1 116 MeV/c2
P46.66
2.52 × 103 K
P46.68
(a) Z0 boson
(b) microwave
(b) 17.6 Gyr
(c) baryon number
(b) gluon or photon